6 Sep 2005 [From Steuard Jensen]
>>> Will the light beam still bounce back to the receptor...
>>> 1. During acceleratiion?
>> No.
...
>> Does all that make sense?
> Yes, it confirms what I thought the answer was.
Excellent. (But you may want to revisit the full explanation once
you've figured out why the answer is "Yes" after the acceleration
stops! I've found that in physics, it can sometimes be all too easy
to come up with the right answer for the wrong reasons.)
>>> 2. At the end of acceleration (once again in uniform motion but
>>> without deceleration)?
>> Yes, the laser will hit the mirror and bounce back to the receptor.
> Oops. Here is one point where I have a problem. My brain is thinking
> that even though the space ship is once again in constant speed, it
> is going faster than it was before the acceleration took place.
> Previously, the beam was aligned to hit the top mirror at the point
> where the mirror would be by the time the beam got to the mirror. If
> the mirror has not been adjusted for the new, faster speed, won't the
> alignment be off at the new, faster speed, since the beam is aligned
> to hit the mirror at the point the mirror would have been at the
> slower speed.
>
> I know that is convoluted language, but do you get my concern??
Absolutely, and it's a legitimate concern! (And the language
wasn't too painfully convoluted, either. :) ) I'll explain at
greater length below.
> In an earlier portion of your last answer you stated: "Then what he
> saw at that point was that the laser was carefully aligned to aim at
> where the mirror _would be_ when the light got there (much as a
> football quarterback throws a pass to where he expects the running
> receiver to end up)." If the rocket ship is now going faster,
> wouldn't the beam have to be adjusted to lead the receiver by a
> bigger margin before we would get a bounce back?
Yes, you're on precisely the right track. But I think I'll include
a couple more quotes from our earlier messages before going into
detail; one of your comments later on seems to have pinpointed
exactly where your puzzlement is coming from.
>> From the point of view of the observer on the rocket, the
>> "gravitational field" has now gone away. Thus, the laser's
>> previous alignment is no longer being disturbed, and the light will
>> bounce off of the mirror just as it did before.
>> From the point of view of the observer who was in an inertial frame
>> the whole time, the laser is once again aiming at just the right
>> angle ahead so that it will hit the mirror. That angle has to be
>> different, of course, since the rocket is now moving at a different
>> speed than it was before.
> Okay, the above statement seems to confirm that at the new faster
> constant speed the beam would not hit the mirror unless and until the
> angle of the beam were adjusted to account for the faster speed. Is
> this interpretation of what you said correct??
Yes, but the "adjustment" in this case happens automatically. (In
particular, note that the observer in the rocket doesn't see any
need to make an adjustment at all!) Just one more excerpt from
your message, and I'll explain in detail.
[From a bit later in your response:]
> Still on this same point, in case I have misunderstood you, I
> understand a basketball inheriting the speed of the train. However, I
> thought light was special and did not inherit the speed of the
> emitter. Am I getting in deeper or am I starting to see the light? I
> can't tell.
Excellent! You've identified the precise point of your confusion,
and it's a very sensible point to be confused. :) With a little
luck, I'll be able to clear up this mystery for you.
The short answer is that in relativity, light _does_ inherit the
speed of the emitter, but in a rather odd way. The "oddness" is
what makes sure that it always ends up moving at the speed of
light. In particular, if the emitter is moving in the same
direction that the light is emitted, that "oddness" means that in
this special case, the light _won't_ inherit any of the emitter's
motion. That's the exception, not the rule.
But that's painfully vague. To actually understand what's going
on, what you need are some details. Now, as is often the case in
physics, there are two ways to work out those details. There's the
easy way that works great _if_ (and only if) you already completely
trust the theory. And there's the brute force way where you just
plug through the math and read off the results.
In this case, the easy way requires that you already trust that
relativity works as advertised. If you do, then my description
above as seen by the observer on the rocket "proves" that the light
must hit the mirror: that observer felt like she was standing still
when she aligned the laser and the mirror, so if she once again
feels like she's standing still, she'll see no reason to realign
the mirror. We know that physical questions like "did the light
hit the mirror" must have the same answer to every observer. So
_if_ we trust relativity, we can conclude that the observer on
Earth _must_ see the angle of the laser beam change after the ship
speeds up, just enough so that it hits the mirror. No math
required! :)
But that sort of explanation isn't at all satisfying when you're
still trying to convince yourself that the theory makes sense.
What you really need is to work through the details entirely from
the point of view of the observer on Earth. I don't know if you've
gotten at all comfortable with the mathematics of relativity yet,
but with luck I can give you a sense of what's going on without
needing to get too specific. But, hmm. How comfortable are you
with calculus? The explanation might work better if I could use it
freely. :) We'll see...
You already know that the observer on Earth sees the clocks on the
rocket running slow, and that the faster the rocket moves, the
slower its clocks seem to run. So here's a strange concept that I
want you to think about: if 't' is the time as the Earth observer
measures it, and 'T' is the time as the rocket observer measures
it, then we can talk about what the Earth observer might call "the
rocket's speed through time". In calculus language, that would be
"dt/dT". Less formally, "dt/dT" just means "how many seconds of
Earth time 't' pass for each second of rocket time 'T'?" For a
rocket going 60% of the speed of light, we find that "dt/dT =
1.25". (I may have given the wrong numerical answer in an earlier
email, now that I think of it. [I've corrected it on this site.]) If
the rocket sent out a flash once per second by its clock, the observer
on Earth would see one flash every second and a quarter.
So far so good. Now, we can also ask what velocities people on the
rocket will measure for various things. In particular, we can ask
what velocity they measure for the laser beam. If 'X' is parallel to
the direction that the Earth observer sees the rocket moving and 'y'
is the direction of the laser beam, then on the rocket they will
measure "v_X = 0" and "v_y = c". (As usual, 'c' is the speed of
light.) In the language of calculus, velocity can be written "v_X =
dX/dT", "v_y = dy/dT": that is, how far in 'X' or 'y' does the light
move for each unit of rocket time 'T'?
Ok. So here's the trick. The observer on Earth will measure the 'y'
velocity as "dy/dt", not "dy/dT"! At least in this simple case, these
can be related by "dy/dt = (dy/dT) / (dt/dT)". (I'm not talking about
the 'x' velocity yet, as it's a bit more complicated. Bear with me.)
So because "dt/dT" is greater than one, "dy/dt" must be less than 'c'.
In particular, for a rocket moving at 60% of 'c', I said that "dt/dT =
1.25", so we'll find that 'dy/dt = 0.8 c' (rather than 'dy/dT = c'
that we had before).
Because light always moves at total speed 'c', that means that the
light's velocity as seen from Earth must have a component in
another direction as well. The only natural other direction is
'x', since that's the direction of the rocket's motion. And what
must the speed in the x direction be? We can use the Pythagorean
theorem (A^2 + B^2 = C^2): we know the "hypotenuse" must be the
total speed 'c'. We know that the 'y' leg has speed "4/5 c". So
as A^2 = C^2 - B^2, we have
(v_x)^2 = c^2 - (4/5 c)^2 = c^2 (1 - 16/25) = c^2 (9/25)
And thus, the 'x' velocity v_x is just "3/5 c": that's 60% of the
speed of light, exactly the same as the rocket's velocity! Thus,
in this case the light _does_ inherit the emitter's velocity.
(This was something of a special case, because the emitter's
velocity was perpendicular to the direction of the light in the
emitter's rest frame. In a more general situation, things get more
complicated.)
To find that answer directly, you would need to relate the coordinate
'X' on the rocket to the (parallel) coordinate 'x' on the Earth.
There's a factor of 1.25 from "length contraction", and there's a
factor of 0.6 because the two coordinate systems are in relative
motion. After dividing by "dt/dT = 1.25", we get 60% of 'c'. But I
won't go into that in detail unless you ask (and unless you're
comfortable with vectors and matrix multiplication!).
At any rate, we've only done this for one specific speed. But you
can see that the answers that we found for the 'x' and 'y'
velocities of the laser beam both depended on the speed of the
rocket relative to Earth. A different speed would have led to a
different angle. And none of that depended on any adjustments to
the mirror: the speed was inherited from the emitter, albeit in a
rather subtle way.
Did that help you to understand this point at all better? I know
it's complicated stuff, so don't hesitate to ask for clarification.
>> Let me explain what each of the three observers sees, given your
>> premise that one observer (on Earth, say) sees each ship moving at
>> 60% of the speed of light in opposite directions.
>> From the point of view of the emitting ship, the other ship is not
>> moving away faster than the speed of light. (In fact, when you
>> work through the math, people on the emitting ship will see the
>> other one moving at about 88% of the speed of light.) So they
>> certainly expect to see the light catch up with the other ship
>> eventually.
> Why is this? If they are receding from each other at 120% the speed
> of light, why does ship A think ship B is moving away at 88% of the
> speed of light.
It's a combination of two effects. First, the clock on ship A is
"running slow" as compared to the clock of the observer on Earth who
measured the 120%. And second, there is a "length contraction"
difference in the way that the two observers will measure distance in
the direction of motion. The details take a bit of computing (some
reasonably messy math and logic, in fact), but hopefully it's not too
surprising that when two observers can't agree on their stopwatches or
on their yardsticks, they won't have much luck agreeing on speeds,
either. :)
> I guess that is the central question I am hung up on. Does it
> inherit the motion of the emitter, and if so, how, why and to what
> degree? Certainly not to the same degree as a bullet from a gun or
> basketball being tossed, does it?
Hopefully I've gotten a good start on answering this above. We've now
talked about two special cases. As seen from the emitter's reference
frame:
* If the light is aimed _perpendicular_ to the relative motion of the
second observer, then the second observer will see the light inherit
all of the emitter's motion. As we saw, the second observer will
see the light's velocity in the "original" direction correspondingly
reduced, just enough for the total speed to remain 'c'.
* If the light is aimed _parallel_ to the relative motion of the
second observer, then the second observer will not see the light
inherit any of the emitter's motion. The light will continue in the
same direction at the same speed, 'c'. (There's no reason that it
would change direction, and it can't move at any other total speed.)
If we were to consider a more general case, in which the relative
angle was somewhere between perpendicular and parallel, the result
would be somewhere between these results as well. The math gets more
complicated, unfortunately, but the "easy" argument will still work:
because every observer must agree about "physical" questions, we know
for sure that the observer on Earth will see the laser move in just
the right direction for it to hit the mirror.
I hope _some_ of that made sense, anyway! I have a feeling that it
got pretty brain-bending at times, so please do ask for clarification
anywhere you need it.