Question on "Relativity of Simultaneity"
Harold Ellis Ensle
In article <350AF1...@sprintmail.com>,
Andrew Biggs <abi...@sprintmail.com> wrote:
>I have just read a passage in a book on relativity (R.W. Lawson's
>translation of Einstein's "Relativity") which has me confusued. I am
>not a physicist nor do I have any experts with whom I can consult on
>this. Could someone please give me a hand with this one?
>
>The section I am refering to is entitled, "The Relativity of
>Simultaneity". In it the author discusses a scenario to illustrate the
>relative nature of "simultaneous" events. Here are the facts of the
>scenario (although you may already familiar with them):
>
>1) A train is traveling along a straight track with velocity w > 0.
>
>2) There are two observers: Observer K1 is on the train and is seated at
>the exact midpoint between the front and rear of the train. Observer
>K2 is standing along the embankment such that at time t=0 he is also at
>the train's midpoint.
>
>3) At time t=0 the train is struck by lightning at the front and rear.
>
>4) The whole system is in vacuo
>
>The passage asserts that observer K2, since he is not "moving", will
>judge the two events at having occurred simultaneously, while observer
>K1 will judge the lightning to have struck the front of the train first.
>
>Now I'll try to explain why this bothers me. The assertion seems to be
>implying that if the train is moving, then a photon originating at the
>front of the train at time t=0 will arrive at the middle of the train
>before a photon originating at the rear of the train at time t=0.
>
>But why should this be? As an observer on the train, my perspective is
>that I am stationary and the earth is rotating beneath me. Why should
>the train's velocity relative to the embankment have anything at all to
>do with whether or not I observe these two event as being simultaneous?
>
>Thanks!
>Andrew Biggs
>
Another mind raped by relativistic reasoning!
Andrew Biggs
Pete Smith
I'm no expert either, but I think I know this one...
The train's motion relative to the embankment is only of relavence because
the embankment happens to be
stationary relative to the *bolts of lightening*. (ie: embankment and
lightening are in the same frame of reference)
If the bolts of lightening happened to be travelling through some medium
which was somehow in motion
relative to the embankment, then it would be the velocity of the train
relative to this medium which would
determine the observed times of arrival of the two lightening bolts.
Indeed, if this 'moving medium' happened to be travelling in the same
direction as the train, at the same speed, then it would be the observer on
the train who would observe the two events as being simultaneous, and the
observer on the embankment who would perceive the bolts arriving at
different times.
I probably haven't worded that very well, but I hope you see what I'm
getting at!
Pete.
Andrew Biggs <abi...@sprintmail.com> wrote in article
<350AF1...@sprintmail.com>...
aart heijboer
Pete Smith wrote:
> I'm no expert either, but I think I know this one...
>
> The train's motion relative to the embankment is only of relavence
> because
> the embankment happens to be
> stationary relative to the *bolts of lightening*. (ie: embankment and
> lightening are in the same frame of reference)
>
> If the bolts of lightening happened to be travelling through some
> medium
> which was somehow in motion
> relative to the embankment, then it would be the velocity of the train
>
> relative to this medium which would
> determine the observed times of arrival of the two lightening bolts.
>
> Indeed, if this 'moving medium' happened to be travelling in the same
> direction as the train, at the same speed, then it would be the
> observer on
> the train who would observe the two events as being simultaneous, and
> the
> observer on the embankment who would perceive the bolts arriving at
> different times.
I can't realy follow this... and call me paranoid, but (unless you
believethese ether-freaks) I get scared when people start talking about
light (or bolts
of lightening, whatever you mean by this) that have a speed relative to
a medium.
Ok, so the answer (I think) is... The observer K2 is a special one,
because to himthe two events (lightening at the front and back) occured
simulaniously.
Now I have no idea about why this should be the case, but given the fact
that it is..
it follows that an observer on the train will observe that the events
dont occur simulatiously.
This is the whole point of these train-stories: K1 and K2 dont agree on
what events are simultanious.
It could just as well have been the case that according to K1 the events
where simultanious
and then they wouldnt have been according to K2.
----------------------------------------------------------------------------------------
I had just typed the story below, when I reread your question and
suddenly understood it....
But no way I'm gonna delete this peace of genius ;-)
Here's the standard train-story for you:
and here's the artists impression:
* M *
| A |
|_________________________| ------------> train moves this
way fast.
OO OO
~~~~B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The * represent light bulbs. A (who is on the train) and
B(on the groud) are people observing these lights.
M is the middle of the train car.
Ok here goes: Suppose that at some moment in time A
observes both bulbs to be flashing at the same time.
This means that A will say the light from both flashes reaches
the middle of the carriage M simultatiously (after all, to A all is
the same as if the train was standing still)
You are of course right when you say that what A sees has nothing
to do with what mr. B happens to be doing.
But now consider mr. B's point of view of the same physical
event (being the light reaching M simultaniously.... there can be
no dissagreement on that fact... you can put a little man in the middle
that shouts "yes" (or some more ammusing phrase) if they do, and both
A and B will either hear 'yes' or they wont)
So the light will reach M simulatiously, but this means that
the left bulb must have flashed before the right one did,
because while the light was moving towards the middle of
the train, the middle itself was moving to the right; meaning
the light from the left bulb just had a larger distance to travel
and because the speed of light is still c, it has taken a longer
time.
concluding A and B do not agree wether the lights flashed simultaniously
even though they are observing the same event(s).
ps: The case of a train with one light in the middle and the
two events (who's simultaniouity (?) the obsevers dissagrea
on) are the light reaching the ends of the carriage is a little
more uhhh natural. But I figured this reassembles the question
better. (by the way I don't think you can have lightening in a vacuum)
Pete Smith
Ok, I think I made a bit of a pig's ear of explaining this before!
The point I was trying to make is this: (And it's a simple point really)
The 'train' observer and the 'embankment' observer are moving, relative to
each other. (They are in different
frames of reference)
Therefore, they observe the two events as having occured at different
times. That's all there really is to it.
I think what's confusing Andrew is this idea of a 'time t=0', which doesn't
really mean anything.
We are really talking about 4 different 'times':
t1 : The time at which the 'train' observer sees the bolt arrive at
the front of the train
t2 : The time at which the 'train' observer sees the bolt arrive at
the back of the train
t1' : The time at which the 'embankment' observer sees the bolt arrive
at the front of the train
t2' : The time at which the 'embankment' observer sees the bolt arrive
at the back of the train
The relationship between t1 and t1', (and t2 and t2' ), is described by the
Lorentz transformations.
It just so happens that in this instance, t1' and t2' happen to be equal.
(Because the light from the front and back of the train arrive at his eyes
at the same time)
This doesn't mean that the 'embankment' observer is 'right' and that the
bolts really *did* occur at the same
time. Nor does it mean that the 'train' observer is right. We could have
100 different observers, travelling at
different relative speeds, who all had different observations about the
arrival times of the bolts, and noone could
say that their observed times were more 'right' than anyone elses.
(Hope this makes sense!)
Pete.
aart heijboer <t...@nikhef.nl> wrote in article
<350B0C90...@nikhef.nl>...
Cees Roos
> I have just read a passage in a book on relativity (R.W. Lawson's
> translation of Einstein's "Relativity") which has me confusued. I am
> not a physicist nor do I have any experts with whom I can consult on
> this. Could someone please give me a hand with this one?
>
> The section I am refering to is entitled, "The Relativity of
> Simultaneity". In it the author discusses a scenario to illustrate the
> relative nature of "simultaneous" events. Here are the facts of the
> scenario (although you may already familiar with them):
>
> 1) A train is traveling along a straight track with velocity w > 0.
>
> 2) There are two observers: Observer K1 is on the train and is seated at
> the exact midpoint between the front and rear of the train. Observer
> K2 is standing along the embankment such that at time t=0 he is also at
> the train's midpoint.
>
you are assuming too much.
E. said: '*if* we assume that in the embankment frame the bolts are
simultaneous, *then* the point where both light flashes will meet in
the embankment frame will be halfway between the points where the bolts
hit in the embankment frame.
Only then does he make the observation, that at the time of the (assumed
simultaneous) strikes, as observed in the embankment frame, both
midpoints, in the embankment frame and in the train frame, must coincide.
So, *if* K2 happens to stand at that point, *then* yes, at that moment in his
frame, he will also stand at the train's midpoint.
Summarizing:
Observer K2 is standing along the embankment midway between the two
lightning bolts, which hit simultaneously (as observed in his frame),
and as a consequence of the premisses, this coincides with the train's
midpoint, at time t = 0 (t=0 is the time in the embankment frame, at
which, in that frame, both bolts hit, as per assumption).
> 3) At time t=0 the train is struck by lightning at the front and rear.
>
At time t=0, as observed in the embankment frame. (Remember: 'if', 'then').
> 4) The whole system is in vacuo
>
> The passage asserts that observer K2, since he is not "moving",
>
Not because he is not moving, but because he is standing midway
between the two lightning strikes, and observes that the lightflashes
meet each other where he is standing.
> will
> judge the two events at having occurred simultaneously, while observer
> K1 will judge the lightning to have struck the front of the train first.
>
> Now I'll try to explain why this bothers me. The assertion seems to be
> implying that if the train is moving, then a photon originating at the
> front of the train at time t=0 will arrive at the middle of the train
> before a photon originating at the rear of the train at time t=0.
>
No. The whole point of the chapter is to show that:
*if* the strikes are observed to be simultaneous in the embankment frame
(because they are equidistant from observer K2, and the flashes reach his
location simultaneously),
*then* they cannot be observed to be simultaneous in the train frame
(because they are equidistant from observer K1, but the flashes do not
reach his location simultaneously).
Summary:
*If* the two bolts are in the embankemnt frame observed to both hapen at
time t=0, *then* they must, as a consequence of the assumption of invariant
c, in the train frame be observed to be happening at two different times
t'.
[snip]
--
Regards, Cees Roos
I think it's much more interesting to live not knowing, than
to have answers which might be wrong. Richard Feynman 1981
Wayne Throop
:: Andrew Biggs <abi...@sprintmail.com>
:: As an observer on the train, my perspective is that I am stationary
:: and the earth is rotating beneath me. Why should the train's
:: velocity relative to the embankment have anything at all to do with
:: whether or not I observe these two event as being simultaneous?
This may sound cryptic and koan-like, but it is actually very
simple and straightforward: think about it for a while.
As an observer in London, my perspective is that I am standing vertical
and the earth curves away from horizontal in all directions. Why should
London's angle relative to New York have anything at all to do with
whether or not I observe two points in Manhattan as being exactly
horizontal from each other?
The answer for the London/NewYork/horizontal case is, because in London
I set my notion of "horizontal" differently from the New York notion of
horizontal, along planes perpendicular to our respective plumb bobs.
Thus, as a Londoner, New Yorkers are leaning at an angle, and
Manhattan streets are all on a steep slope.
Similarly for the train/embankment/simultaneous case; because on the
train, I set my notion of "simultaneous" differently from the embankment
notion, along hyperplanes that make light propogation isotropic in our
respective coordinate systems (that is, the similarity is making light
propogation isotropic in space and time, vs making gravity gradient
isotropic in space; it's how you make a time axis of a coordinate system
"at right angles" or "perpendicular" to spatial offsets).
So. In the case you're talking about, the two events are GIVEN as
simultaneous on the embankment, just as we might be GIVEN two points exactly
horizontal from each other in New York. And thus, the train's velocity
relative to the embankment "has to do" with whether the events are
simultaneous in train coordinates, exactly the same thing as what
the London/NewYork relative angle WRT the earth's center has to do
with whether the given points in manhattan are exactly horizontal
in London coordinates.
: Harold Ellis Ensle <hen...@ix.netcom.com>
: Another mind raped by relativistic reasoning!
Translating from the Ensle-ish, and given the contexts where he
has said similar things in the past, this seems to mean,
"another mind that can deal with simple geometry".
( Mind you, I do NOT suspect Ensle is incapable of dealing with simple
geometry. He merely seems to refuse to do so in the case of
SR/Einstein/Minkowski, for motives completely unknown to me. )
--
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Rbwinn
>
>1) A train is traveling along a straight track with velocity w > 0.
>
>2) There are two observers: Observer K1 is on the train and is seated at
>the exact midpoint between the front and rear of the train. Observer
>K2 is standing along the embankment such that at time t=0 he is also at
>the train's midpoint.
>
>
>
>judge the two events at having occurred simultaneously, while observer
>K1 will judge the lightning to have struck the front of the train first.
>
>Now I'll try to explain why this bothers me. The assertion seems to be
>implying that if the train is moving, then a photon originating at the
>front of the train at time t=0 will arrive at the middle of the train
>before a photon originating at the rear of the train at time t=0.
>
>that I am stationary and the earth is rotating beneath me. Why should
>the train's velocity relative to the embankment have anything at all to
>do with whether or not I observe these two event as being simultaneous?
Andrew,
I am not a scientist or mathematician either, and this identical
explanation caused me to become interested in what scientists were saying.
Scientists will give you the explanation that the train is shorter in the
direction that the train is moving because of a distance contraction and that
the two bolts of lightning strike at different times in the frame of reference
of the train. This explanation is, of course, nonsense. However, by using
this convoluted reasoning, scientists are able to get their mathematics to do
what they want it to do.
My analysis of the problem was this. Lightning strikes the front oand rear
of the train, and we stipulate that they are simultaneous in the frame of
reference of the train. What does each observer see?
The time each flash of light will take top reach the observer on the train
is equal to half the length of the train divided by the speed of light. The
observer will see both flashes of light at the same instant.
If each bolt of lightning leaves a mark on the railroad track, then an
observer on the ground who was opposite the observer on the train when the
lightning struck will be midway between the two marks on the railroad track.
The distance between the marks on the track will be the length of the train,
and the light from each bolt of lightning will reach the observer on the ground
at a time of 1/2 the length of the train divided by the speed of light in his
frame of reference.
Remember that at the time the observer on the train sees the light, the
train has traveled a distance down the track, and also, that when the observer
on the ground sees the light , the train has traveled a distance down the
track. I developed a set of equations showing this relationship that
scientists are not happy about that show that the speed of the train is
different when measured from on the train than when measured from on the
ground.
Robert B. Winn
and...@ibm.net
The point of this experiment has nothing to do with scaling
lengths and times in transforming from one frame to another.
It basically shows that any reasonable transformation has
to mix the time and space coordinates together to produce
the new time coordinate if all observers measure the same
speed of light.
Robert Winn's equations do not include this and therefore
don't predict that all observers measure the same speed of
light except along the forward direction of motion.
If Winn responds to this, I will just repost the Robert Winn
is Wrong FAQ which explicitly shows this.
John Anderson
Edward Schaefer
Andrew Biggs wrote:
>
> I have just read a passage in a book on relativity (R.W. Lawson's
> translation of Einstein's "Relativity") which has me confusued. I am
> not a physicist nor do I have any experts with whom I can consult on
> this. Could someone please give me a hand with this one?
>
> The section I am refering to is entitled, "The Relativity of
> Simultaneity". In it the author discusses a scenario to illustrate the
> relative nature of "simultaneous" events. Here are the facts of the
> scenario (although you may already familiar with them):
>
>
> 2) There are two observers: Observer K1 is on the train and is seated at
> the exact midpoint between the front and rear of the train. Observer
> K2 is standing along the embankment such that at time t=0 he is also at
> the train's midpoint.
>
> 3) At time t=0 the train is struck by lightning at the front and rear.
>
> 4) The whole system is in vacuo
>
> The passage asserts that observer K2, since he is not "moving", will
> judge the two events at having occurred simultaneously, while observer
> K1 will judge the lightning to have struck the front of the train first.
>
> Now I'll try to explain why this bothers me. The assertion seems to be
> implying that if the train is moving, then a photon originating at the
> front of the train at time t=0 will arrive at the middle of the train
> before a photon originating at the rear of the train at time t=0.
That is exactly what happens. To see why this is the case, let us
look at the case of a train that is 600m long. In this case,
observer K2 see two flashes of lightning occur 300m away at t=1 ms.
Since 1 ms - [(300 m)/c] = 0, he knows they both occurred at t=0.
For observer K1, the situation is different. Since he has been
moving during the 1 ms it took the light to get to K2, he is now
closer to the source of the lighting at the front of the train
in the frame-of-reference of K2, and is at (300 m) * (c-v) / c
from its source when the light gets to him. At the same time,
he is now, in K2's frame, (300 m) * (c+v)/c from the source of
the lightning from the back of the train. Therefore, K1 sees the
light from the front stroke first and soon afterwards sees the
light from the rear stroke.
This much is simple Newtonian physics. It should not bother you
much.
In relativity, Observers K1 and K2 must agree on the following
points:
- That the lightning strokes hit at the front and rear of the
train.
- That K1 is located in the center of the train (and was beside
K2 at t = t' = 0).
- That the speed of light wrt themself is c.
Since K1 did not have the flashes of light from equidistant
positions reach him at the same time, they cannot have occurred
simultaneously. OTOH, they must have been simultaneous for K2
becuase the flashes DID reach him from equidistant positions
at once.
EMS
Rbwinn
>
>If Winn responds to this, I will just repost the Robert Winn
>is Wrong FAQ which explicitly shows this.
>
>John Anderson
>
>
John has appointed himself as the protector ofdistance contraction. My
equations are:
sq root (x'^2 + y'^2 + z'^2) = sq root(x^2 + y^2 +z^2) -vt
t'=t(c-v)/c
vt=proper speed*t'
I will briefly explain why John's proof does not apply to these equations.
John uses cosines as they are used with the Lorentz equations to show inertial
relationships between two frames of reference. My equations do not use
inertial relationships, but are calculated using only distances and times
without any distance contraction or relativity of simultaniety.
Consequently, if light is emitted in a moving frame of reference S' when the
origin of S' is at the origin of S a system at rest, the light emitted moves
out from the origin of S at a rate of c, and therefore will be seen at any
point in S according to a sphere with a radius of ct with its center at the
origin of S.
Since the point where the light was emitted in S' moves with the moving
frame of reference, light in S' will be represented by a sphere with a radius
of ct' with its center at the origin of S'. The origin of S' will be a
distance of vt from the origin of S.
John's proof makes light a physical material rather than energy.
Robert B. Winn
and...@ibm.net
The following is a mini-FAQ on the posts of Robert Winn.
It is not meant to convince Winn, as he can't seem to follow
simple algebraic arguments. He has been shown several
times that his equations are inconsistent and just resorts
to incomprehensible handwaving to avoid dealing with it.
This document just serves to provide a concise rebuttal
of Winn's posts for the benefit of other readers of the
group. I intend to post this FAQ every day that Winn posts
one of his little gems. I do not intend to respond to
any response that Winn makes to these postings.
Two reference frames S and S'. S' is moving with velocity
v in the positive x direction with respect to S. The coordinates
t = t' = 0, x = x' = 0, y = y' = 0 describe the same event.
I have suppressed the z coordinate since it isn't important to
the discussion. At the event (0,0,0) (coordinates in either
frame, a light is flashed so that a sphere of light spreads
out from this event. Since I suppressed the z coordinate,
the light spreads out in a ring in the x-y plane.
S frame
=========================================================
light sphere spreads uniformly from the spatial
origin starting at t = 0
At time t
x = ct cos(O)
y = ct sin(O)
where O is the angle between the particular direction
and the x axis
y
. .
. .
. .
. .
. . O
................................... x
.
.
.
.
.
.
.
.
The sin and cos of any angle satisfy
(cos(O))^2 + (sin(O))^2 = 1
So you see that x^2 + y^2 = (ct)^2 for any direction O.
Also cos(0) = 1, cos(90) = 0, cos(180) = -1,
sin(0) = 0, and sin(90) = 1
where the angles are measured in degrees.
S' frame
==========================================================
In this frame, the light sphere is spreads out unifromly
from the spatial origin starting at t' = 0
Similarly to the S frame, you can write
x' = ct' cos(O')
y' = ct' sin(O')
The above only really assumes that the light sphere is
centred on the spatial origin of both frames and starts
at time 0 in both frames and that light travels in straight
lines.
Now transform the S coordinates using the Lorentz transformation
and Winn's equations.
Lorentz
======================================================
g = (1 - (v/c)^2)^-0.5
t' = g(t - vx/c^2) = g(t - tv cos(O)/c)
x' = g(x - vt) = g(ct cos(O) - vt)
y' = ct sin(O)
But x' = ct' cos(O') and y' = ct' sin(O') so
cos(O') = (cos(O) - v/c) / (1 - v cos(O)/c)
sin(O') = sin(O)/g(1 - v cos(O)/c)
There are two things to note here:
First
(cos(O'))^2 + (sin(O'))^2 = 1
To see this,
(cos(O) - v/c)^2 + (sin(O))^2/g^2
= (cos(O))^2 - 2 cos(O) v/c + (v/c)^2 + (1 - (v/c)^2) (sin(O))^2
= (cos(O))^2 + (sin(O))^2 - 2 cos(O) v/c + (v/c)^2 (1 - (sin(O))^2)
= 1 - 2 cos(O) v/c + (v/c)^2 (cos(O))^2
= (1 v cos(O)/c)^2
This shows that, contrary to Winn's oft repeated assertion, the
light spreads out spherically in both frames according to Lorentz.
Second, the direction perpendicular to the direction of motion
in the S' frame is given by cos(O') = 0, but for cos(O') = 0,
cos(O) = v/c. In other words, there is light emitted in the
perpendicular direction in the S', it's the light that is
emitted at O = arccos(v/c) in the S frame.
This shows that, contrary to Winn's oft repeated assertion,
the S' observer at 90 degrees to the direction of motion
will see light according to Lorentz.
Winn
==============================================================
t' = t(1 - v/c)
x' = x - vt
y' = ? (since Winn includes y and y' in his eqautions
but never specifies how to transform y)
But we don't need y' to see what's wrong with these equations.
t' = t(1 - v/c)
x' = ct cos(O) - vt
x' should be ct' cos(O') so
cos(O') = (cos(O) - v/c) / (1 - v/c)
But if the angle O' is a real angle, you require
-1 <= cos(O') <= 1
But cos(O') >= -1 if
(cos(O) - v/c)/(1 - v/c) >= -1
=> cos(O) - v/c >= -1 + v/c
=> cos(O) >= 2v/c - 1
But this means that the angle O can't take on all of the
values between 0 and 180 degrees. There is some range
of O for which the absolute value of cos(O') > 1,
which means O' can't be a real angle and there's no
way you can compensate for this by fiddling with the
definition of y' because if |cos(O')| > 1, there is
no way that (cos(O')^2 + (sin(O'))^2 = 1.
This shows that Winn's equations are inconsistent
with his own description of how light propagates.
Also note, cos(O') = 0 for cos(O) = v/c, the same
as for Lorentz, so that for the perpendicular direction
in the S' frame, Winn says the light came from the
same direction as Lorentz in the S frame. So whatever
Winn thinks is wrong with Lorentz in the perpendicular
direction is also wrong with Winn.
John Anderson
ESTAPA
There are two fixed sources of light between which I am driving my car at
100 mph. I am driving away from one source on an exact straight line and
exactly toward the second source. I symultanioulsy measure the light speed
of the following light and the oncoming light using the same watch in the
same instant along the same distance inside the car and both results are
c.???
Hypostasis:
Could light be traveling at all speeds but only visible in a c frame?
Could I be measuring a part of light traveling 100 MPH faster from the rear
and a slower part of light traveling 100 MPH slower from the front while
both yield c when measured inside the car?
Could an observer traveling away from a light pulse see the pulse sooner
than a stationary observer (with respect to the source) when the pulse
reaches the moving observer and at the same instant both observers are the
same distance from the source?
D. Estapa
Rbwinn
>Could I be measuring a part of light traveling 100 MPH faster from the rear
>and a slower part of light traveling 100 MPH slower from the front while
>both yield c when measured inside the car?
>Could an observer traveling away from a light pulse see the pulse sooner
>than a stationary observer (with respect to the source) when the pulse
>reaches the moving observer and at the same instant both observers are the
>same distance from the source?
>D. Estapa
>
>
D Estapa,
Light is energy which reacts with the elements of the universe at a rate
of c. The only frame of reference which has any real meaning with regard to
light is the frame of reference in which it is being measured. There is no way
to prove anything about light in any other frame of reference with present
levels of knowledge.
Robert B. Winn