# A scientist’s transmitter emits a wavelength

1. Mar 19, 2007

### 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?

2. Mar 19, 2007

### MeJennifer

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

3. Mar 19, 2007

### pervect

Staff Emeritus
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).

4. Mar 19, 2007

### Staff: Mentor

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

5. Mar 19, 2007

### 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.

6. Mar 19, 2007

### 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.

7. Mar 19, 2007

### 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?

8. Mar 19, 2007

### 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.

Last edited: Mar 19, 2007
9. Mar 19, 2007

### 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].

Last edited: Mar 19, 2007
10. Mar 19, 2007

### 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.

Last edited: Mar 19, 2007
11. Mar 19, 2007

### 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].

12. Mar 19, 2007

### 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.

Last edited: Mar 19, 2007
13. Mar 19, 2007

### 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.

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14. Mar 19, 2007

### pmb_phy

Why is that?

Pete

15. Mar 19, 2007

### 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.

16. Mar 20, 2007

### 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

17. Mar 20, 2007

### 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:
$$\] \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} \[$$

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.

18. Mar 20, 2007

### MeJennifer

Robphy, any particular reason the line is sloped with respect to OT.
Let's make it a bit simpler, assume that OT and EF are at relative rest with each other.

So let's for the sake of argument assume that you are right and that the emission and absorption of the photon do not happen at the same time:

So let's clock synchronize, using the Einstein clock synchronization method OT frame.
Suppose E is emitted at 10:20 in the morning at what time is it absorbed at R?

With regards to the distance between the approaching photon and the event of absorbtion:

Pick any point you like on the photon's worldline and call it point A.
Then take the point where the photon and the wordline of the observer crosses and call this B.

Now calculate the distance between point A and B.

The distance between those two points is zero, under all circumstances!

Last edited: Mar 20, 2007
19. Mar 20, 2007

### robphy

Yes, I agree that the spacetime interval is zero between those two events A and B. Great job.

But many of the other comments that you made in this thread made little sense... using the standard terminology of special relativity and Minkowski spacetime. In particular, this makes no sense using standard terms:
That quoted statement is simply false. I offered a correction to replace "observer's plane of simultaneity" with "the past light cone of the absorption event".

20. Mar 20, 2007

### MeJennifer

So then answer what the time difference is between the emittance and absorption of the photon for an Einstein clock synchronized frame of reference.
I say it is zero, what do you say?

Let's take a "practical" example, say we put clocks everywhere between us and some object X which is 5 light hours away from us, which, for symplicity's sake, is at rest relative to us.
We synchronize all clocks using Einstein's method.
Then say at 10:30 in the morning we emit a photon in the direction of the object X. What will be the time, on the local clock near the object, when the photon is absorbed?

Last edited: Mar 20, 2007
21. Mar 20, 2007

### pmb_phy

MJ - I posted a few questions for you in my last post in this thread. However you never answered the questions. Perhaps you thought that they were rehtorical? Let it be known that they were not rehtorical questions. I made some assumptions about what you seemed to be trying to say. Those assumptions were made so that I might arrive at the same conclusion that you did. But how can I be sure of this if you don't answer??

Best regards

Pete

22. Mar 20, 2007

### MeJennifer

Sorry for that.

Ok, you asked me when I wrote:

what I mean by "conform the Minkowski metric".

Well, the metric of space-time is such that the distance between any two points on a lightlike worldline is always zero. This in contrast to for instance an Euclidean metric in Galilean space-time.

Then you wrote:

Yes, and the distance between any point on the photon's worldline and the absorbtion event is zero.

23. Mar 21, 2007

### ramcg1

I seemed to have germinated a discussion (worth while of course) in a direction other than what I originally intended. I neglected to mention Special Relativity.

I asked this question because I wanted to find out how special relativity handles energy changes under the Relativistic Doppler effect. The energy has to come from somewhere or go somewhere yet all inertial frames of reference are equal, c is the same and different observers will measure the frequency of the photon and obtain different results according to their relative velocity with respect to the source.

If you want you could imagine scientist One emitting a stream of long wave radiation photons and scientist Two detecting these as high energy gamma rays due to the relativistic Doppler effect – a massive increase in energy. If you want to put in some form of independent collision to cause the gamma ray to form the electron positron pair; fine – go ahead. The question is essentially the same. In SR observers are supposed to have god like powers of observation so scientist One should be able to deduce that it is his original photon that has undergone the transformations.

24. Mar 22, 2007

### pervect

Staff Emeritus
Special relativity handles energy much the same way that Newtonian physics does. (I'm going to avoid talking about how GR handles energy, it's unfortunately more complex). But the case you are talking about is very simple and very similar to Newtonian theory.

In any given inertial frame, the total energy of an isolated system is a constant.

However, this does not imply that a "boost" transformation (a transformation from one inertial frame to a different inertial frame that's moving relative to the intitial frame) does not change the energy of a system.

In fact, it should be obvious that even in Newtonian physics, a "boost" of a system changes its energy. If you have a baseball that is standing still, if you perform a "boost" transformation to a different frame, the baseball now has kinetic energy - energy that it did not have before.

However, this energy did not "come from" anywhere. The energy of a system depends on the frame of reference used to measure it.

Your question simply replaces the baseball with a gamma ray. Just as in the baseball case, the energy of an isolated system in any given inertial frame is a constant. Also as in the baseball case, when you perform a "boost" on the system, the energy of the system will not be the same as it was for the "unboosted" system.

Last edited: Mar 22, 2007
25. Mar 22, 2007

### ramcg1

"The energy of a system depends on the frame of reference used to measure it."

Are you implying that by changing the frame of reference you change the total energy of a system? How does tha tconserve energy.