A scientist’s transmitter emits a wavelength

ramcg1

A scientist’s transmitter emits a wavelength of very long wave electromagnetic radiation. We call him One and he watches his wave move through the universe at c.

Another scientist, call him Two, approaches One at very nearly c.
Two moves directly at One’s wave and nearly dies of fright when he sees a very high energy gamma ray approaching. Fearing possible cell damage and malignant cancer he turns on his magnetic deflector.

Just before the gamma ray reaches the magnetic deflector it spontaneously disintegrates into an electron / positron pair. The design of the magnetic deflector is such that the electron and positron are separated in the vacuum of pace and now moving in a plane perpendicular to the direction back to One.

One is scratching his head trying to work out how his very long wave turned into an electron / positron pair.

Later the electron and positron collide and the resultant gamma ray returns towards a dismayed One.

Where did the extra energy come from?

MeJennifer

That is impossible.
The space-time interval between an approaching photon and an observer is always zero.

pervect

Hint: Model the process as a collision (see nits). What happens to Two's velocity after the collision with the gamma ray?

Various nits: Two wouldn't see the gamma ray coming, magnetic fields don't deflect gamma rays (they would deflect electrons or positrons), single photons in free space can't decay into an electron/positron pair unless they collide with something (energy and momentum must be conserved, which is not possible unless there is something else for the photon to interact with).

jtbell

How does Two "see" the gamma ray approaching, before it meets him?

robphy

I think you mean to say that the interval between the emission event and reception event of the photon by the observer is zero. Intervals are between events, not (say) worldlines.

Of course, this is a technical way to say that there is no way to "react" to an approaching gamma ray since nothing [in particular, information about the emission event] outruns light.

MeJennifer

Right.

But note that as soon as a photon was emitted in the direction of the observer then in the observer's frame of reference, using Einstein synchronized clocks, it has already been absorbed.

robphy

Are you saying that the target observer determines the emission event to be simultaneous [i.e. assigns the same time coordinate t (in accordance with his wristwatch)] with the absorption event?

MeJennifer

If we assume the observer uses Einstein clock synchronization then the answer is yes. Both the emission and the absorption of the photon are on the observer's plane of simultaneity.

However if the observer is accelerating away it could possibly outrun the absorbtion of the photon, but it would need to accelerate forever.

robphy

Hopefully, you realize that, in a nice spacetime like Minkowski spacetime, two distinct events that are simultaneous according to some observer are spacelike-related, whereas two distinct events on a photon's worldline are lightlike-related.

What is true is that both the emission and absorption events are on the past light cone of the absorption event [where the photon meets the observer].

MeJennifer

Yes I realize that.
Is there anything I wrote that might contadict that?

Some would possibly say that issue seems to be a bit beyond the scope of run of the mill relativity.
But I would love to see, in a new topic, some proof of that assertion.

robphy

Yes.
Distinct emission and absorption events of a photon can't both be spacelike related [if they are simultaneous for some observer] and lightlike related [if they are events of a photon].

MeJennifer

The sub-lightspeed particle that emits the photon is indeed spacelike related to the observer, however that is not the case for the emitted photon. On the plane of simultaneity you will notice that both the temporal distance and spatial distance between the emitted photon and the observer is zero.

robphy

Take a look at Ellis and Williams "Flat and Curved Spacetimes" p. 48
to see the discussion of the past light cone vs the spacelike plane of simultaneity ("world map")... in ordinary Minkowski spacetime.

[by request, attachments added]

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pmb_phy

Why is that?

Pete

MeJennifer

Because, conform the Minkowski metric, the length of the photon's wordline is always zero, so then it follows that the distance between each point on this line and the space-time event that constitutes the absorbtion must be zero as well.

pmb_phy

What do you mean by "conform the Minkowski metric"??

What you said whas this
A photon lies on a null geodesic and the spacetime interval between all photons on the same null geodesic is zero. So what you're saying is that the photon is on the null geodesic and that the observers worldline crosses the geodesic at the point of intersection between photon and observer. Hmmm. Yeah. That makes sense. Thanks MJ.

Pete

robphy

I think you're either confused or using standard terms in nonstandard ways.

The terms "spacelike" or "timelike" or "lightlike"-related only refer to a pair-of-events, not a pair-of-observers (or a pair-of-observer-worldlines).

Here is a [Minkowski] spacetime diagram of the situation you describe:
[tex]
\]
\begin{picture}(200,200)(0,0)
\unitlength 2mm
{
\qbezier(0,50)(0,0)(0,0)\put(0,0){O}
\qbezier(0,0)(60,0)(60,0)\put(0,50){T}
}
{
\put(30,50){S}
\qbezier(30,50)(40,0)(40,0)
\qbezier(00,10)(40,0)(40,0)
\put(0,10){F}
}
{
\put(40,0){E}
\qbezier(40,0)(0,40)(0,40)
\put(0,40){R}
}
{
\qbezier(0,0)(40,40)(40,40)
\put(33,33){P}
\put(20,20){X}
}
\end{picture}
\[
[/tex]

ES is the [timelike] worldline of the sub-lightspeed particle (the source)
EF is that particle's [spacelike] plane of simultaneity [which is Minkowski-orthogonal to ES]

OT is the [timelike] worldline of the observer (the receiver/absorber)
OE is the observer's [spacelike] plane of simultaneity [which is Minkowski-orthogonal to OT]

E is the emission event by the source
R is the reception/absorption event by the observer
ER is a lightlike spacetime path. In fact, ER is on the past-light-cone of event R.

Note that E and R are not simultaneous for the observer OT or the particle ES.
In fact, since E and R are lightlike-related, E and R cannot be spacelike-related, and thus cannot be simultaneous events for any timelike observer. That is why I commented on your earlier statements:

Now let's parse your more recent post:
As I said above, spacelike-related only refers to pairs-of-events, not pairs-of-worldlines. While O-and-E and F-and-E are spacelike-related pairs of events (one on each worldline), O-and-P and E-and-R are lightlike-related and O-and-S and E-and-T are timelike-related.
Here, one can say that every event on the photon's spacetime path ER is lightlike-related to event R on the observer's worldline. However, note that the events on the segment XR are timelike-related to O on the observer's worldline, and the events on the segment EX are spacelike-related to O.

So, in summary, it is incorrect to use the term "spacelike-related" with anything but pairs-of-events.

For the OT observer, the planes parallel to OE are his [spacelike] planes of simultaneity.
For each event Y on ER excluding event-R itself, the OT-observer determines the spatial-distance between event-Y and his worldline OT to be nonzero.

The only way I can see to make a true statement with a subset of your words is this... paraphrasing... The distance-between-these-two-lines OT and ER is zero since they intersect. But that will be true of any two intersecting lines, regardless of the nature of their tangent vectors.