Re: relativity: what if you can measure how far the light traveled?
- From: bennett@xxxxxxxxxxxxx
- Date: 12 Feb 2007 01:16:21 -0800
On Feb 11, 3:58 pm, "Jim Black" <trams...@xxxxxxxxx> wrote:
benn...@xxxxxxxxxxxxx wrote:
Newbie relativity question:
My book says the Principle of Relativity says that any observer can
assume they are stationary and the usual laws such as the constant
speed of light will still apply. But I'm having trouble with this
apparent paradox:
Suppose person A is at km marker 0, and B is at km marker 1. A has
two laser guns pointed toward B except they are separated by a tiny
angle such that after traveling 1 km, the beams will be exact 1 meter
apart.
A fires the guns, and at that moment B is still stationary. Then B
immediately starts moving toward km marker 2 at half the speed of
light. Since B can move 1 km in the time the beams can travel 2 km,
they arrive there at the same time. Since the beams traveled 2 km
they are now 2 meters apart.
I take it you're making the approximation here that the component of
the laser beams' velocity along the line from A to B is the speed of
light. It's actually a little less than the speed of light, because
the laser beams are at a small angle from that line. Specifically, if
we let f be the slope of the laser beams, 1/2 meters/kilometer, then
the velocity component along the line is c/sqrt(1+f^2). There's
nothing wrong with making this approximation; just be aware that we're
making it, and don't be surprised when we have to drop a factor of
sqrt(1+f^2) at some point.
But, switch to B's frame of reference. From that POV, A fired two
lasers toward B from 1 km away, and then took off in the opposite
direction while B stayed still. But light continues moving in the
same direction no matter what the source does afterwards, so from B's
point of view, the beams would travel 1 km, and when they arrive at B
they should be 1 meter apart!
First, A is not one kilometer away when the beam is fired. Here's the
spacetime diagram in A's reference frame:
[View diagram in a fixed-width font such as Courier. You may have to
copy the text into a word processor to change the font.]
(1,2)
| /
| //
| / /
| / /
| / /
| / /
| / /
| / /
Person A | Laser Beams / / Person B
| / /
| / /
| / /
| / /
| / /
| / /
| / /
| / /
| / /
| / /
| / /
| / /
| / /
| / /
|/______________________/
(-1,0) (0,0)
Above, I'm using units of kilometers and light-kilometers, and I'm
putting the 1-km marker at x=0 to make the transformation easier.
After a Lorentz transformation to B's frame of reference, we have:
(0,2/g)
/|
/ |
/ |
/ |
\ Person A Laser Beams / | Person B
\ / |
\ / |
\ / |
\ / |
\ / |
\ / |
\ / |
\ / |
\ / |
(-g,-g/2) \-__ |
\ --__ |
\ --__ |
\ --__ |
\ --__ |
\ --__ |
\ --__|
(0,0)
Above, g is the relativistic gamma factor, 1/sqrt(v^2/c^2), which in
this case is equal to sqrt(4/3).
So far, we have the light traversing a distance of sqrt(4/3) km, so it
looks like the beams will be sqrt(4/3) meters apart when they hit B.
But we're not done yet. Remember that we transform to B's frame of
reference, A will undergo length contraction:
A's Frame of Reference
_
_/
_/ _/
/ _/
o _/
-|A|-
|_| _
/ \ \_
\_ \_
\_ \
\_
B's Frame of Reference
/
/ /
/ /
o /
-|A|-
|_|
/ \ \
\ \
\ \
\
If we assumed that the laser beams traveled in the same direction the
lasers were pointed, this would multiply the final distance between
the laser beams by another factor of gamma. That would give us a
final separation of 4/3 meters between the beams.
But that assumption would be wrong -- the laser beams don't travel in
the same direction as the lasers are pointed, because the laser is
moving. In the diagrams below, a light pulse is leaving a moving
laser:
+---------------+
| / |
| / / |
| / / |
| / / |
| / / |
| * / |
+---------------+
+---------------+
| / |
| / / |
| / / |
| / / |
| / * / |
| / |
+---------------+
+---------------+
| / |
| / / |
| / / |
| / * / |
| / / |
| / |
+---------------+
+---------------+
| / |
| / / |
| / * / |
| / / |
| / / |
| / |
+---------------+
+---------------+
| / |
| / * / |
| / / |
| / / |
| / / |
| / |
+---------------+
+---------------+
| / * |
| / / |
| / / |
| / / |
| / / |
| / |
+---------------+
Notice that the path of the light is different from the orientation of
the laser:
+---------------+
| * |
| * |
| * |
| * |
| * |
| * |
+---------------+
To calculate the direction the light will go, we draw a vector
diagram:
*/
ct * /
* /
* /
* /
*---->/
vt
But for our problem, Person A is traveling in the negative x
direction, so the diagram should look like this:
_
_/*
_/ *
_/ * ct
_/ *
_/ *
/<----*
vt
Remember we said f was the slope of the laser beams. Let's let g be
the slope of the lasers. We mark up our diagram in preparation for a
calculation:
- . . . . . ._
| _/*.
_/ * .
y _/ * ct
_/ * .
| _/ * .
- /<----* .
. vt . .
. . .
. |-----|
. y/f .
. .
|--- y/g ---|
We have:
sqrt(y^2/f^2 + y^2) / (y/g - y/f) = (ct) / (vt) = c/v
Multiplying the numerator and denominator by f/y:
sqrt(1 + f^2) / (f/g - 1) = c/v
Dropping that factor of sqrt(1 + f^2):
1 / (f/g - 1) = c/v
f/g = 1 + v/c = 3/2
This last correction factor brings the distance between the beams when
they strike Person B up to 2 meters.
--
Jim E. Black- Hide quoted text -
- Show quoted text -
Jim,
Thank you very, very much for your thorough answer. This seems to
make the most sense, although I'm still working on trying to resolve
what seem to be apparently paradoxes, but are probably caused by a
faulty assumption on my part somewhere... More posts probably to
follow.
.
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