"Randy Poe" wrote in message
ups.com...

Androcles wrote:
"Randy Poe" wrote in message
ups.com...

Notice on this journey from earth to mars that mars
is moving toward the spacecraft.

Huh? You sure of that?

No. I deleted the message a couple of minutes after
posting it. I misread that particular diagram, in which
the rocket is playing catchup with Mars.

Thinking about this a little more, I realize that this
does make sense since the orbital velocity you start with
from Earth is faster than that of Mars.

Actually (further reading on "Hohman transfer") it appears
that the Hohman transfer orbit is in the SAME direction as
orbits of earth and mars in BOTH directions, and
if I understand it what you're doing is putting yourself
into an ellipse around the sun which closely matches
the velocities of both earth and mars. The orbital mechanics
do the velocity matching for you. On the way home, you
catch up with earth's speed just from the KE you gain
by falling toward the sun.

Very good. Now you can appreciate why a launch window is
important. Mars and Earth must be points on the ellipse at the
appropriate times.



This is all from considerations of minimum energy. I
was thinking of minimum time, where I'm pretty sure
that you'd want to go TOWARD the body you're trying
to reach.

Actually it is a little more complex than that. You need to
match velocities as well, or you'll have a disaster from
which there is no recovery. You can blast away from
the Earth radially from the Sun, but you'll still retain
Earth's tangential velocity of 30,000 km/sec.
If you give that up and head backwards to meet Mars
coming from behind, not only will you be wasting a
huge amount of KE but you'll impact as well. So now you
have to burn more fuel to gain the same tangential
velocity of Mars.



Wasn't Joe walking away
from the mosquito, yet the distance between them
still closed?

Yes. The question is why you'd want to play catch
up instead of going the other way. The answer is that
this minimizes the energy you have to expend.

In either direction, the description of a rocket moving
while the planet sits still is not a very useful
description.

- Randy
Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?
Androcles.

Androcles wrote:
"Randy Poe" wrote in message
ups.com...

This is all from considerations of minimum energy. I
was thinking of minimum time, where I'm pretty sure
that you'd want to go TOWARD the body you're trying
to reach.

Actually it is a little more complex than that. You need to
match velocities as well, or you'll have a disaster from
which there is no recovery.

The recovery is to adjust your velocity before
landing. I think a slingshot maneuver can help you
save fuel in either gaining or shedding KE.

You can blast away from
the Earth radially from the Sun, but you'll still retain
Earth's tangential velocity of 30,000 km/sec.
If you give that up and head backwards to meet Mars
coming from behind, not only will you be wasting a
huge amount of KE but you'll impact as well. So now you
have to burn more fuel to gain the same tangential
velocity of Mars.

But a well-designed slingshot might help you do
most of that adjustment without burning fuel.

Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?

All observers will measure the speed of light
to be 299792458 m/sec exactly.

- Randy

Androcles wrote:

Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?
Androcles.




When you say "relative to the vehicle", do you mean the speed of a
photon as measured in the vehicle's own rest frame, or do you mean the
speed that the distance between the vehicle and a photon grows, as seen
in the sun's rest frame? You understand that relativity only says the
first one has to be equal to c, not the second one, correct?

Jesse

"Randy Poe" wrote in message
ups.com...

Androcles wrote:
"Randy Poe" wrote in message
ups.com...

This is all from considerations of minimum energy. I
was thinking of minimum time, where I'm pretty sure
that you'd want to go TOWARD the body you're trying
to reach.

Actually it is a little more complex than that. You need to
match velocities as well, or you'll have a disaster from
which there is no recovery.

The recovery is to adjust your velocity before
landing. I think a slingshot maneuver can help you
save fuel in either gaining or shedding KE.

You can blast away from
the Earth radially from the Sun, but you'll still retain
Earth's tangential velocity of 30,000 km/sec.
If you give that up and head backwards to meet Mars
coming from behind, not only will you be wasting a
huge amount of KE but you'll impact as well. So now you
have to burn more fuel to gain the same tangential
velocity of Mars.

But a well-designed slingshot might help you do
most of that adjustment without burning fuel.

Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?

All observers will measure the speed of light
to be 299792458 m/sec exactly.


So you assert but cannot prove. Tell me how they will do it.

"Jesse Mazer" wrote in message
...


Androcles wrote:

Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?
Androcles.



When you say "relative to the vehicle", do you mean the speed of a
photon as measured in the vehicle's own rest frame,

Yes.

or do you mean the speed that the distance between the vehicle and a
photon grows, as seen in the sun's rest frame? You understand that
relativity only says the first one has to be equal to c, not the
second one, correct?

Jesse
I understand that Einstein STARTED with
"light is always propagated in empty space with a definite velocity c
which is independent of the state of motion of the emitting body"
and then attempted some invalid mathematics, yes.
I also undertand that he used c+v and c-v in his computation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
and said "But the ray moves relatively to the initial point of k, when
measured in the stationary system, with the velocity c-v, so that
x'/(c-v) = t."

When I went to school, c-v was not equal to c.
It still isn't.
Androcles.

Androcles wrote:

"Randy Poe" wrote in message
oups.com...


Androcles wrote:


"Randy Poe" wrote in message
groups.com...


This is all from considerations of minimum energy. I
was thinking of minimum time, where I'm pretty sure
that you'd want to go TOWARD the body you're trying
to reach.


Actually it is a little more complex than that. You need to
match velocities as well, or you'll have a disaster from
which there is no recovery.


The recovery is to adjust your velocity before
landing. I think a slingshot maneuver can help you
save fuel in either gaining or shedding KE.



You can blast away from
the Earth radially from the Sun, but you'll still retain
Earth's tangential velocity of 30,000 km/sec.
If you give that up and head backwards to meet Mars
coming from behind, not only will you be wasting a
huge amount of KE but you'll impact as well. So now you
have to burn more fuel to gain the same tangential
velocity of Mars.


But a well-designed slingshot might help you do
most of that adjustment without burning fuel.



Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?


All observers will measure the speed of light
to be 299792458 m/sec exactly.




So you assert but cannot prove. Tell me how they will do it.





To understand why the speed of light is the same in all reference
frames, you first must understand the physical definition of a
"reference frame" in relativity. A reference frame is really just a
coordinate system for marking the location and time of different events.
You can think of the physical basis of these coordinates in terms of
each observer having a network of rulers and clocks spread throughout
space, so if you imagine taking a picture of an event as it happens, its
coordinates are defined by the markings on the ruler and the time on the
clock right next to the event in the photo. Each observer uses a network
of rulers and clocks which are at rest relative to himself. Clocks at
different locations must all be synchronized, and in classical physics
this would be pretty simple--you could just bring two clocks together to
check that they're in sync, then move them to different positions. But
time dilation due to movement makes sychronization a little trickier.
Einstein's idea was that you could have each observer synchronize their
own set of clocks using light signals--pick a point halfway between two
clocks, send a flash of light from that point in both directions, and if
both clocks read the same time when the light reaches them, they are
considered synchronized. But unlike the Galilean frames used in
classical physics, this means that the different Lorentzian frames used
in relativity will not agree about simultaneity of pairs of events. For
example, if an observer on a spaceship sees a flash of light go off in
the middle of the ship, then he should synchronize clocks at the front
and back of the ship by making sure they both read the same time when
the light hits them. But for another observer who sees the ship moving,
the back of the ship is moving *towards* the point where the flash
happened and the front is moving *away* from that point, so his own
network of clocks will be synchronized in such a way that the clock next
to the event "light hits the back of the ship" will read an earlier time
than the clock next to the event "light hits the front of the ship".

I drew some diagrams a while ago to help explain how different reference
frames could each see the other's rulers contracted and the other's
times dilated without leading to any contradictions. In this example, we
have two rulers with clocks mounted on them moving alongside each other,
and in order to make the math work out neatly, the relative velocity of
the two rulers is (square root of 3)/2 * light speed, or about 259.628
meters per microsecond. This means that each ruler will observe the
other one’s clocks tick exactly half as fast as their own, and will see
the other ruler's distance-markings to be squashed by a factor of two.
Also, I have drawn the markings on the rulers at intervals of 173.085
meters apart—the reason for this is again just to make things work out
neatly, it will mean that observers on each ruler will see the other
ruler moving at 1.5 markings/microsecond relative to themselves, and
that an observer on one ruler will see clocks on the other ruler that
are this distance apart (as measured by his own ruler) to be out-of-sync
by exactly 1 microsecond, some more nice round numbers.

Given all this, here is how the situation would look at 0 microseconds,
1 microsecond, and 2 microseconds, in the frame of ruler A:

http://www.jessemazer.com/images/RulerAFrame.gif

And here’s how the situation would look at 0 microseconds, 1
microsecond, and 2 microseconds, in the frame of ruler B:

http://www.jessemazer.com/images/RulerBFrame.gif

Some things to notice in these diagrams:

1. in each ruler's frame, it is at rest while the other ruler is moving
sideways at 259.6 meters/microsecond (ruler A sees ruler B moving to the
right, while ruler B sees ruler A moving to the left)

2. In each ruler's frame, its own clocks are all synchronized, but the
other ruler's clocks are all out-of-sync

3. In each ruler's frame, each individual clock on the other ruler ticks
at half the normal rate. For example, in the diagram of ruler A’s frame,
look at the clock with the green hand on the -519.3 meter mark on ruler
B--this clock first reads 1.5 microseconds, then 2 microseconds, then
2.5 microseconds. Likewise, in the diagram of ruler B’s frame, look at
the clock with the green hand on the 519.3 meter mark on ruler A—this
clock also goes from 1.5 microseconds to 2 microseconds to 2.5 microseconds.

4. Despite these differences, they always agree on which events on their
own ruler coincide in time and location with which events on the other.
If you have a particular clock at a particular location on one ruler
showing a particular time, then if you look at the clock right next to
it on the other ruler at that moment, you will get the same answer to
what that other clock reads and what marking it’s on regardless of which
frame you’re using. Here’s one example:

http://www.jessemazer.com/images/MatchingClocks.gif

Formally, the equations for transforming between two different
Lorentzian reference frames S and S', where S' is moving at velocity v
relative to S along its x-axis (and S is moving at velocity -v relative
to S' along its x' axis) would be:

x'=gamma(x - vt)
y'=y
z'=z
t'=gamma(t - vx/c^2)

x=gamma(x' + vt')
y=y'
z=z'
t=gamma(t + vx'/c^2)

where gamma = 1/squareroot(1-v^2/c^2)

These equations describe how readings on one frame's rulers and clocks
will match up with readings on another frame's rulers and clocks,
assuming each set of clocks is synchronized by light signals as I
described earlier. You can use these transformations to derive the fact
that each frame sees the other's rulers contracted by a factor of
squareroot(1-v^2/c^2), the fact that each frame sees the other's clocks
tick at 1/squareroot(1-v^2/c^2) times slower than the correct rate, that
velocities as measured in different frames are related by the equation
(u + v)/(1 + uv/c^2), and so on.

Compare with the equations for transforming between the Galilean
reference frames used in Newtonian physics:

x'=x - vt
y'=y
z'=z
t'=t

x=x' + vt
y=y'
z=z'
t=t'

Even if Newtonian physics were true, if you had a network of rulers and
clocks and used them to label the coordinates of an arbitrary event
(x,y,z,t), you could easily use the Lorentz transformation equations to
find the same event's coordinates in a different Lorentzian reference
frame (x',y',z',t'), and you would still find that anything traveling at
c in the first coordinate system was also traveling at c in the second.
So you can see that the statement "c moves at the same speed in all
reference frames" is not really a physical statement at all, but just a
mathematical consequence of using the Lorentz transformations to define
the coordinate systems of different reference frames! But if Newtonian
physics were true, these other frames would no longer have physical
meaning in terms of the readings of actual rulers and clocks because
there'd be no length contraction or time dilation, and because
synchronizing clocks by light-signals wouldn't make sense if you could
synchronize clocks by bringing them together and then moving them apart.
Also, if Newtonian physics were true, the laws of physics would not
always have the same form when stated in terms of (x',y',z',t')
coordinates as they would when stated in terms of (x,y,z,t) coordinates.
For example, Newtonian gravity is assumed to be an instantaneous force,
so if an object changes its path the gravitational effects will be felt
instantaneously by distant objects; but two spatially separated events
with the same t-coordinate won't have the same t'-coordinates, so you
would have to use different equations to describe how gravity works in
terms of these coordinates.

But what is actually seen in modern physics is that all our most
accurate laws of physics are Lorentz-invariant--they have exactly the
same form when stated in different Lorentzian coordinate systems. So
although the statement "c moves at the same speed in all reference
frames" is just a mathematical consequence of using the Lorentz
transformations to define different reference frames, and would be true
regardless of what the laws of physics were like as long as we defined
reference frames this way, the physical content of relativity lies in
the fact that the laws of physics are Lorentz-invariant, which is a
nontrivial observation about how things work in our universe as opposed
to some other possible universe (like one where Newtonian physics is
precisely accurate), and it explains the practical value of defining
reference frames in the Lorentzian way as opposed to the Galilean way.

As for the question of *why* all our most accurate laws of physics
exhibit Lorentz-invariance, I don't think there's any answer to that.
It's similar to the question of why the laws of physics exhibit
translation-invariance (if you write the equations in one coordinate
system, they will have the same form when written in another coordinate
system where the spatial origin is displaced relative to the first
coordinate system's origin) or why they exhibit rotation-invariance (if
you write the equations in one coordinate system, they will have the
same form when written in another coordinate system whose spatial axes
are rotated relative to the first one's).

Jesse

Androcles wrote:
I understand that Einstein STARTED with
"light is always propagated in empty space with a definite velocity c

which is independent of the state of motion of the emitting body"
and then attempted some invalid mathematics, yes.
I also undertand that he used c+v and c-v in his computation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] =
tau(x',0,0,t+x'/(c-v))

And does the form x'/(c-v), or x'/(c-v) + x'/(c+v)
look familiar?

Did you perhaps see anything like that when you worked
out the mosquito problem?

Didn't you write this?

t1 = 32/(5-3) = 16

And this?

t2 = [ 16 + 32/(5+3) ] = 20.

Did you write them so long ago that you can no longer
remember what those denominators mean?

and said "But the ray moves relatively to the initial point of k,
when
measured in the stationary system, with the velocity c-v, so that
x'/(c-v) = t."

When I went to school, c-v was not equal to c.

And the mosquito is moving, according to a stationary
observer, at 5-3 = 2 m/sec. Yet we started out assuming
the mosquito moves at 5 m/sec, and 5 is not 2.

How, oh how, can we reconcile the 5 with the 2? When I
went to school, 5 was not 2.

Could it POSSIBLY be that "the speed in the stationary
system" and "the speed relative to a moving object, as
measured in the stationary system" are NOT THE SAME THING?
- Randy

"Randy Poe" wrote in message
oups.com...

Androcles wrote:
I understand that Einstein STARTED with
"light is always propagated in empty space with a definite velocity c

which is independent of the state of motion of the emitting body"
and then attempted some invalid mathematics, yes.
I also undertand that he used c+v and c-v in his computation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] =
tau(x',0,0,t+x'/(c-v))

And does the form x'/(c-v), or x'/(c-v) + x'/(c+v)
look familiar?

Sure, and unequal.


Did you perhaps see anything like that when you worked
out the mosquito problem?

Of course. Everyone knows (1/2) * (16+4) = 16.
That's how you prove the velocity of a mosquito is 5 fps
in all inertial frames of reference.
You are a relativist, so you are bound to agree, right?



Didn't you write this?

t1 = 32/(5-3) = 16

Yes, I did.


And this?


t2 = [ 16 + 32/(5+3) ] = 20.

Yes.
I also wrote (well, copied it from Einstein, actually)

½[tau(0,0,0,t)+tau(0,0,0,t+16+4)] = tau(32,0,0,t+16)

In good faith, of course, which you do not seem to have much of,
as you are trying to save face now. The monkey has won, Poe.
Now I'll drool sarcasm if you keep up your idiocy. The question is,
which one is the monkey?



Did you write them so long ago that you can no longer
remember what those denominators mean?

Good grief no. In good faith they mean 5-3 and 5+3.

and said "But the ray moves relatively to the initial point of k,
when
measured in the stationary system, with the velocity c-v, so that
x'/(c-v) = t."

When I went to school, c-v was not equal to c.

And the mosquito is moving, according to a stationary
observer, at 5-3 = 2 m/sec. Yet we started out assuming
the mosquito moves at 5 m/sec, and 5 is not 2.

Ah, but you don't see, the DISTANCE measured by the stationary
observer isn't 64 feet, because 20 seconds (agreed?) * 2 fps = 40 ft
The mosquito is moving at 5 fps in all inertial frames of reference,
but only records 64/5 = 12.2 seconds by his own watch. It's called
time dilation, I'm sure you've heard of it (in good faith).




How, oh how, can we reconcile the 5 with the 2? When I
went to school, 5 was not 2.

Time dilation, of course. Simple, Isn't it?


Could it POSSIBLY be that "the speed in the stationary
system" and "the speed relative to a moving object, as
measured in the stationary system" are NOT THE SAME THING?

Good grief no, the speed of mosquitoes is 5 fps in all inertial frames
of reference. Einstein's mathemagics proves it.

Androcles.

"Jesse Mazer" wrote in message
...


Androcles wrote:

"Randy Poe" wrote in message
roups.com...

Androcles wrote:

"Randy Poe" wrote in message
egroups.com...

This is all from considerations of minimum energy. I
was thinking of minimum time, where I'm pretty sure
that you'd want to go TOWARD the body you're trying
to reach.

Actually it is a little more complex than that. You need to
match velocities as well, or you'll have a disaster from
which there is no recovery.

The recovery is to adjust your velocity before
landing. I think a slingshot maneuver can help you
save fuel in either gaining or shedding KE.


You can blast away from
the Earth radially from the Sun, but you'll still retain
Earth's tangential velocity of 30,000 km/sec.
If you give that up and head backwards to meet Mars
coming from behind, not only will you be wasting a
huge amount of KE but you'll impact as well. So now you
have to burn more fuel to gain the same tangential
velocity of Mars.

But a well-designed slingshot might help you do
most of that adjustment without burning fuel.


Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?

All observers will measure the speed of light
to be 299792458 m/sec exactly.



So you assert but cannot prove. Tell me how they will do it.




To understand why the speed of light is the same in all reference
frames, you first must understand the physical definition of a
"reference frame" in relativity.

Do me a favour.
Einstein wrote

"light is always propagated in empty space with a definite velocity c
which is independent of the state of motion of the emitting body"
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Then he wrote:
"But the ray moves relatively to the initial point of k, when measured
in the stationary system, with the velocity c-v, so that x'/(c-v) = t."
and
"It follows, further, that the velocity of light c cannot be altered by
composition with a velocity less than that of light. For this case we
obtain V = (c+w)/(1+w/c) = c."
but he wrote that last AFTER he had "composed" c-v and c+v.

[snip wasted argument]
Androcles.

Androcles wrote:

"Jesse Mazer" wrote in message
...


Androcles wrote:



"Randy Poe" wrote in message
groups.com...



Androcles wrote:



"Randy Poe" wrote in message
legroups.com...



This is all from considerations of minimum energy. I
was thinking of minimum time, where I'm pretty sure
that you'd want to go TOWARD the body you're trying
to reach.



Actually it is a little more complex than that. You need to
match velocities as well, or you'll have a disaster from
which there is no recovery.



The recovery is to adjust your velocity before
landing. I think a slingshot maneuver can help you
save fuel in either gaining or shedding KE.




You can blast away from
the Earth radially from the Sun, but you'll still retain
Earth's tangential velocity of 30,000 km/sec.
If you give that up and head backwards to meet Mars
coming from behind, not only will you be wasting a
huge amount of KE but you'll impact as well. So now you
have to burn more fuel to gain the same tangential
velocity of Mars.



But a well-designed slingshot might help you do
most of that adjustment without burning fuel.




Yes, the best way to think of the problem is an absolute
frame of reference centred on the sun. What do you think
should be the velocity of light in such a frame?
What should it be relative to the vehicle?



All observers will measure the speed of light
to be 299792458 m/sec exactly.



So you assert but cannot prove. Tell me how they will do it.





To understand why the speed of light is the same in all reference
frames, you first must understand the physical definition of a
"reference frame" in relativity.



Do me a favour.
Einstein wrote

"light is always propagated in empty space with a definite velocity c
which is independent of the state of motion of the emitting body"
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Then he wrote:
"But the ray moves relatively to the initial point of k, when measured
in the stationary system, with the velocity c-v, so that x'/(c-v) = t."


I don't know what the context of this is. I assume he's not talking
about how fast the light is moving in a given frame, but rather how fast
the light is moving away from some other object, as seen not in the
object's own frame but in a frame where the object itself is moving at
velocity v. In this case, although light will still travel at c in this
frame, the distance between the light ray and the object moving at
velocity v will be seen to grow at the rate (c-v) in this frame. In the
object's own frame, though, the distance between itself and the light
ray would grow at the rate c, as relativity predicts.

and
"It follows, further, that the velocity of light c cannot be altered by
composition with a velocity less than that of light. For this case we
obtain V = (c+w)/(1+w/c) = c."
but he wrote that last AFTER he had "composed" c-v and c+v.

[snip wasted argument]
Androcles.







Did you even *read* my argument? Tell me, which of the following
statements do you disagree with?

1. If we use the following equations for transforming between two
Lorentzian reference frames S and S', where S' is moving at velocity v
relative to S along its x-axis (and S is moving at velocity -v relative
to S' along its x' axis):

x'=gamma(x - vt)
y'=y
z'=z
t'=gamma(t - vx/c2)

x=gamma(x' + vt')
y=y'
z=z'
t=gamma(t + vx'/c2)

gamma = 1/squareroot(1-v2/c2)

....then if something has velocity c in frame S, it must also have
velocity c in frame S', according to these equations.

Agree/Disagree?

2. If each observer defines x,y,z, and t coordinates in his own frame by
building a network of rulers and clocks, with the coordinates assigned
to any event to be determined by the readings on the rulers and clocks
in the immediate vicinity of the event, and with all the rulers and
clocks at rest relative to the observer, and with the clocks
synchronized using the *assumption* that light travels at the same speed
in all directions, then the Lorentz transformation equations I gave
above will be the correct ones--if an event happens at coordinates
(x,y,z,t) according to rulers and clocks at rest in frame S, then the
same event will be measured to have the coordinates (x',y',z',t')
according to rulers and clocks in frame S', *if* you use this
synchronization procedure.

Agree/Disagree?

3. If you express some law of physics (say, Maxwell's laws) using
equations that make use of one inertial frame's x,y,z,t coordinates, and
you want to know how the same law looks in another inertial frame which
uses x',y',z',t' coordinates, then as long as the law is
Lorentz-invariant, the equations won't change, all you have to do is
replace x with x', y with y', z with z', and t with t'. All the modern
laws of physics that we know of have this property of Lorentz-invariance.

Agree/Disagree?

4. Therefore, even if there was a single "true" reference frame,
absolutely no physical experiment could distinguish it from any other
reference frame, as long as the laws governing the experiment are
Lorentz-invariant.

Agree/Disagree?