I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?

"Steven" wrote in message ups.com...
I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?


Because space and time are not like you think they are.

Martin Hogbin

Steven wrote:
I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?

In cases like this, you have to be very precise about the term
"relative velocity". You might mean how fast the distance
between the two objects is changing in your frame of reference.
This is referred to as the "closing velocity", and in this case
it would be 1.8c, as you would expect. The other definition is
how fast one object is moving *in the frame* of the the other.
This will always be less than c. If this seems counterintuitive,
just remind yourself that you don't often travel near the speed
of light, so intuition is not of much use.

-jc

I got it!

The distance changes based on the frame of reference.

If from the perspective of the stationary object, the two objects have
just separated by distance d, then from the perspective of either of
the moving objects, the distance has only increased by a fraction of d.
Length constriction makes the distance seem to be much smaller. It's
like saying 10 miles in one frame of reference is only 5 miles in
another.

Does that sound right?

jc wrote:

Steven wrote:
I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?

In cases like this, you have to be very precise about the term
"relative velocity". You might mean how fast the distance
between the two objects is changing in your frame of reference.
This is referred to as the "closing velocity", and in this case
it would be 1.8c, as you would expect.

That's possible, but it's worth noting that in that case we
aren't talking about the velocity of either object in any frame.

The other definition is how fast one object is moving *in the
frame* of the the other. This will always be less than c.

Right. This is the normal definition, as it *is* the velocity of
an object in a frame. To find that velocity, we use the
hyperbolic trigonometry of special relativity.

define: A = arctanh(u/c) B = arctanh(v/c)

V_total/c = tanh(A + B)
= (tanh(A) + tanh(B))/(1 + tanh(A)tanh(B))
= (u/c + v/c)/(1 + uv/c^2)

For u = .9c and v = .9c we have,

V_total/c = (.9 + .9)/(1+.81) ---- V_total = .9945c


--Tim Shuba---

In sci.physics.relativity, Steven

wrote
on 21 Jun 2005 14:34:38 -0700
. com:
I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?


v_O1 = +.9c
v_O2 = -.9c
v_12 = (v_O1 - v_O2) / (1 - v_O1*v_O2/c^2)
= 1.8c / (1 + 0.81) = 180/181 c

where I've used v_xy = velocity of x relative to y.

--
#191,
It's still legal to go .sigless.

Seeing is believing:
Tevatron Luminosity
http://www.fnal.gov/pub/now/tevlum.html
Live Tevatron Status
http://www-bd.fnal.gov/notify/tevsta.../tev_stat.html

Sue...

"Steven" wrote in message
ups.com...
I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?

.................................................. ....
To Steven,

If A travels towards a stationary observer D from the left at .9c and B
travels towards the same stationary observer D from the right at .9c, A and
B will travel towards each other at
.9c + .9c = 1.8c, no matter what the relativists try to tell you to the
contrary.

The observed velocity of a moving body is equal to (vc) / (v + c)
consequently:
D will observe A approach from the left at -.9c / .1 = -9c and B from the
right also at -9c.
(The velocity v is shown negative when the body travels towards the
observer.)

A will not see B travel towards him, and B will not see A travel towards him
because their relative velocity is greater than the speed of light c. This
is similar to a supersonic plane which a listener can't hear coming because
it travels faster than the speed of sound.

Enjoy, Len.

PS, For more information see my website: http://www2.rideau.net/gaasbeek
.................................................. ......

After serious thinking Steven wrote :
I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?

The nature of the question reveals a hidden assumption which are no
longer valid in special relativity.
That assumption is that the group of velocity transformations is
additive (abelian).

Let me give you another example to show the principle and then explain.
Two people standing on the Earth walk 1000 miles. One walks north and
the other south.
Why are they not 2000 miles apart? (straight line distance through the
earth, not over surface).
Or try it with one walking East and one walking North, why are they
not 1414 miles apart?
(i.e. sqrt(2) x 1000 miles) = diagional of 1000x1000 mile square?

The explaination is because the Earth is curved. Thus translations of
position on the surface
do not add like vectors the way they do on a flat plane. Rather you
must view translations
of positions in terms of rotations about the center of the Earth.

The group for motions on the surface of a plane is:
ISO(2)
(Inhomogenous Special Orthogonal group in 2 dim
or the group of 2 rotations and 2 dim translations),

while the motions on the surface of a sphere is SO(3)
(Special Orthogonal group in 3 dimensions or group of 3 rotations
(about center of sphere x^2 + y^2 + z^2 = R^2 ))

Now if you follow that point then consider that for Gallilean
relativity the assumption
is that velocities add (and also can be rotated) so the group of
Gallilean velocity
transformations is ISO(3).

But Special Relativity assumes the surface of velocities is no longer
flat.
It is rather a hyperbaloid (also called a pseudo-sphere)
Vx^2 + Vy^2 + Vz^2 - c^2 Vt= c^2
where the velocities are the rates of change of coordinates with
respect to
proper time, tau (and c = speed of light), and the coordinates
(x,y,z,t) are
those of the moving object relative to a given observer.

The transformation group is now a group called SO(3,1) (the Lorentz
group)
which you may view as "pseudo-rotations" analogous to rotations about
a sphere but involving a hyperboloid of rotation and thus invoking
those funny hyperbolic trig functions you find on scientific
calculators,
sinh, cosh, tanh, etc.

So when you talk about a velocity of 0.9 c you are talking in SR about
a hyperbolic rotation in (x,t) space of about:
theta = 1.47222 "pseudo-radians?"
relative to the observer measuring this velocity.

The velocity dx/dt is the ratio of proper velocity components:
dx/dt = (dx/dtau) / (dt/dtau) = Vx/Vt = c sinh(1.47222)/cosh(1.47222)
= c tanh(1.47222)

The pseudo-rotation (boost) in the x-t plane gives:
Vx = c sinh(theta)
and
Vt = cosh(theta) (the gamma factor)

So let's change your question to "what is the velocity of one of these
..9c
objects relative to the other?" We then instead of adding their
velocities
add the hyperbolic angles of pseudo-rotation (just as we would add
angles
of rotation for people walking in opposite directions on Earth).

You then get a pseudo-angle of 2.94444 which has hyperbolic-tan
0.9946. Thus the two objects see each other moving away at speed
0.9946 c.

In summary the set of object velocities in Special Relativity is no
longer
a three dimensional flat space. It is rather the three dimensional
surface
of a hyperboloid of rotation in four dimensions (x,y,z,t). Thus just
as
on a sphere you do not add finite displacements but rather their angles
of rotation, we likewise cannot add velocities as vectors but rather
must
work out the relative pseudo-rotation angles between two objects'
velocities.


I hope this clearifies more than confuses.

--
Regards,
James Baugh

"Len Gaasenbeek" wrote in message
...

"Steven" wrote in message
ups.com...
I'm sure this question has been asked many times before, but here
goes...

If two objects are moving in opposite directions, and both are moving
at .9c relative to a stationary object, how is it that their speeds
relative to each other do not surpass the speed of light?

.................................................. ...
To Steven,

If A travels towards a stationary observer D from the left at .9c and B
travels towards the same stationary observer D from the right at .9c, A
and
B will travel towards each other at
.9c + .9c = 1.8c, no matter what the relativists try to tell you to the
contrary.

From D's frame of reference that appears to be the case. But not from A's
or B's.

The observed velocity of a moving body is equal to (vc) / (v + c)
consequently:
D will observe A approach from the left at -.9c / .1 = -9c and B from the
right also at -9c.

So, we start with the stationary frame D watching A moving towards it
at -0.9c. And to see what that looks like from the point of view of ... the
stationary frame D watching A moving towards it at -0.9c (sic - yes, the
same frame), we multiply the v by 10.

Breathtaking.

'Til this point no knowledge of relativity or the reams of empirical proof
have been required to see that you've goofed, but now I'll leave the
following to the experts.

A will not see B travel towards him, and B will not see A travel towards
him
because their relative velocity is greater than the speed of light c.
This
is similar to a supersonic plane which a listener can't hear coming
because
it travels faster than the speed of sound.