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Michele
2002-Apr-27, 06:06 AM
Hi, I really liked this site, though I think maybe it goes a bit too easy on the latest Star Trek (Enterprise) /phpBB/images/smiles/icon_smile.gif

I have only an introductory college physics class as my background which pretty much amounts to zero, I'm well aware, though I've read Asimov's Understanding Physics, and have tried to plow through some of Feynman's Lectures. Anyway, Relativity isn't making any sense to me at all (well, duh, it's counterintuitive!), but I wonder if I might gain any tips on this discussion board.

My first obstacle has to do with diagrams used to explain the Michelson/Morey experiment and Lorentz equation. The math is basic trig, but I don't understood why the diagrams give lateral momentum to light beams. (Y'know, those moving train or car diagrams with a photon or light beam being emitted sideways).

For example, if we consider a luminous object moving through space perpendicular to our line of sight (on earth), isn't it true that light that leaves the object and reaches earth (our eyes) will give the image of the light at the location from which it was emitted, instead of being translated by a distance equalling the time it takes for the light to reach our eyes multiplied by the object's lateral speed? What I mean to say is, I thought that the object would not give the emitted light any lateral momentum as a result of the object's transverse motion. Otherwise, when it arrived, the light would indeed indicate the object's true present position instead of its position of, say, 10 minutes ago if the object is 10 light minutes away from us.

I thought this was the case, but this line of thinking must be wrong, since all the diagrams explaining relativity show the light path of a photon being emitted and reflected inside a moving car, for example, as being a diagonal as viewed by an outsider. Disregarding the obvious fact that no observer could see the photon (the photon would have to leave the system and hit the viewer's retina to do that), it is clearly assumed that the photon has lateral momentum and thus travels a diagonal path, instead of what I thought would happen, i.e. that the light would go straight outward regardless of the speed of the vehicle and, if the vehicle were traveling fast enough, the photon would in fact miss the reflecting mirror and so wouldn't come back anyway. And of course if that were the case, all the trigonometric equations would be irrelevant. What's the real story on this, can anyone help?

Thanks,

GrapesOfWrath
2002-Apr-27, 09:05 AM
Point us towards a specific illustraion in Feynman Lectures, and we'll discuss it in particular.

And welcome to the BABB!

<font size=-1>[Welcome!]</font>

<font size=-1>[ This Message was edited by: GrapesOfWrath on 2002-04-27 05:05 ]</font>

2002-Apr-27, 12:15 PM
<a name="20020427.4:10"> page 20020427.4:10 aka & other times2
On 2002-04-27 02:06, Michele wrote: To: pre Math HOUR
8:3:20 A.M. PST {pre/pre/mathhr} see next line
ftp://ftp.dartmouth.edu/pub/gnuplot
6: April 29, 2002 5:13 A.M. Math hour arived
5: not much i would like to say, however
4:
3:
2:
1:
--
1:
2:
3:
well ? i though there would be lots of LInes {oh well} see next post 2

<font size=-1>[ This Message was edited by: HUb' on 2002-04-28 07:26 ]</font>

<font size=-1>[ This Message was edited by: HUb' on 2002-04-29 09:16 ]</font>

2002-Apr-27, 12:19 PM
<a name="20020427.4:14"> page 20020427.4:14 aka I second this motion
On 2002-04-27 05:05, GrapesOfWrath wrote: To: 4:14
Point us towards a specific illustraion in Feynman Lectures, and we'll discuss it in particular.

And welcome to the BABB!

<font size=-1>[Welcome!]</font>
only i would add..
Lorentz equation

John Kierein
2002-Apr-27, 12:52 PM
Michelle. I like you already. Keep up the thoughts. There are lots of folks who don't agree with the idea that there isn't an ether. Any medium sorta acts like an ether when it slows light. There are conflicting ideas of what's happening when light is slowed by a transparent medium. Some, (and this is the prevailing view), say that the light is absorbed and re-emitted by the medium, causing a delay. This should show an ether-like characteistic and light should be deflected in a wind tunnel if this were so. I prefer the view that individual photons are scattered by the medium and the wavefront is reconstructed by Huygens secondary wavelets. The individual photons travel a longer path causing the light to be slowed but not deflecting along with the medium. Interesting in relation to the Michelson-Morley attempts to find the ether wind. The solar wind should deflect light if the first theory is correct. Nowadays we may be able to try to detect this. I predict failure. Good to hear that people are still thinking about this.

Michele
2002-Apr-27, 04:36 PM
Thanks for the replies.

As to diagrams, there are two in the (abridged) Feynman lectures book I have under the lecture on Special Relativity (subsection "Michelson-Morley experiment". He gives a schematic diagram of the experiment where it was attempted to show movement of the earth through an ether, and the diagram has the light beam reflected at the first half-mirror moving in what Feynman calls a zig-zig motion (as the earth moves). Later he gives a different diagram (under subsection "Transformation of time") in relation to a "light clock". The light leaves a flashtube, hits a mirror and bounces back to a photocell. With the transverse motion of the light clock, it is assumed the light takes a diagonal path here too.

And since the time it takes the light to travel the diagonal should be longer than straight out to the mirror and back, given that c is a constant, a discrepancy arises in a relationship based on the square root of a difference of squares, leading to the Lorentz equation.

See you tomorrow,

Michele
2002-Apr-27, 04:53 PM
[quote]
On 2002-04-27 08:52, John Kierein wrote:
Michelle. I like you already. Keep up the thoughts. There are lots of folks who don't agree with the idea that there isn't an ether. Any medium sorta acts like an ether when it slows light. There are conflicting ideas of what's happening when light is

Thanks,
I'm not sure exactly what to think yet on the ether. You might find it funny that while reading Asimov's explanation of the early physicists' understanding of the ether, I was nodding to myself, wow that makes a lot of sense (I was sold) -- and then he goes on to explain how the ether theory was smashed to rubble.... C'est la vie. But in his explanation of the M/M experiment, my impression, perhaps erroneous, was that the experiment didn't prove the lack of an ether per se, but merely the lack of an ether wind. I don't know if that's correct but it's just my reaction. I guess there could still be an ether which only acts as a light propagation medium and has no "wind" relative to any matter? Is that what the modern ether proponents are saying, I wonder?

Cheers,

2002-Apr-28, 11:35 AM
<a name="20020428.3:27"> page 20020428.3:27 aka Li Speed
On 2002-04-27 12:53, Michele wrote: To: Light Speed






? the M/M / ether wind / I wonder?

its the detector responce time
that i have a problem discribing
in Math terms, on old light speed questions. My choice = Omega!

Michele
2002-Apr-29, 11:51 AM
? the M/M / ether wind / I wonder?
[/quote]
its the detector responce time
that i have a problem discribing
in Math terms, on old light speed questions. My choice = Omega!


I'm not exactly sure what you mean, but if it's the light detector you're referring to, the response time at rest is described by 2d/c where d is the distance from the light source to the reflector. But for an outside observer when the vehicle is moving, if the vehicle moves forward a distance 2u, the time the light took to travel was [2*(u^2 + d^2)^1/2]/c, which is a greater amount of time than 2d/c. Hence time dilation. Actually Feynman explains it differently but I think the above way of putting is right.

2002-Apr-29, 01:30 PM
<a name="20020429.5:17"> page 20020429.5:17 aka OMEGA.?
On 2002-04-29 07:51, Michele wrote: To: 2 CHUEN 5 UO
My reply to you seneence structure, indicates to me
that you are currently headed in the direction
of logical Positivism? Well good luck!
i'll begin on OMEGA.. but will run out of lines
as I only get 9.. or so. Omega was {maybe is} a long range
Very Low Frequency radio Loran system. It did operate
around 11KHz or {radio waves at audio frequencies}
as the Earth turned & the Sun Rose & set.. there exists
a shift in the arival time from one place on
Earth to another. making it appear that their distances change ? [out of lines] you get the idea? .A--B. Radio wave / & change? find: EXPLAIN

SeanF
2002-Apr-29, 03:49 PM
Michelle,

Welcome to the Board, and good question! /phpBB/images/smiles/icon_smile.gif

The simple idea is that of inertia. The photon has the emitter's lateral motion before it's emitted, and inertia causes it to maintain that lateral motion.

It's just like with a gun. If the gun and target are both moving laterally at the same velocity, the bullet will still hit the target - it will continue to move laterally along with the gun and target while it is moving between the gun and the target.

If you can prove that the photons don't behave this way, you'll be well on your way to disproving the Theory of Relativity! /phpBB/images/smiles/icon_smile.gif

GrapesOfWrath
2002-Apr-29, 03:55 PM
On 2002-04-27 02:06, Michele wrote:
What I mean to say is, I thought that the object would not give the emitted light any lateral momentum as a result of the object's transverse motion. Otherwise, when it arrived, the light would indeed indicate the object's true present position instead of its position of, say, 10 minutes ago if the object is 10 light minutes away from us.

I thought this was the case, but this line of thinking must be wrong, since all the diagrams explaining relativity show the light path of a photon being emitted and reflected inside a moving car, for example, as being a diagonal as viewed by an outsider. Disregarding the obvious fact that no observer could see the photon (the photon would have to leave the system and hit the viewer's retina to do that), it is clearly assumed that the photon has lateral momentum and thus travels a diagonal path, instead of what I thought would happen, i.e. that the light would go straight outward regardless of the speed of the vehicle and, if the vehicle were traveling fast enough, the photon would in fact miss the reflecting mirror and so wouldn't come back anyway. And of course if that were the case, all the trigonometric equations would be irrelevant. What's the real story on this, can anyone help?

I've given this some thought, as I'm not absolutely sure of what your question is. Are you aware of the phenomenon known as the aberration of light (http://www.harcourt.com/dictionary/def/1/7/3/0/17300.html)? It seems like that may be relevant. Stars emit light in all directions, but the photons that reach us do not arrive from the purely perpendicular direction. The ones that are perpendicular...just miss.



On 2002-04-27 12:36, Michele wrote:
As to diagrams, there are two in the (abridged) Feynman lectures book I have under the lecture on Special Relativity (subsection "Michelson-Morley experiment". He gives a schematic diagram of the experiment where it was attempted to show movement of the earth through an ether, and the diagram has the light beam reflected at the first half-mirror moving in what Feynman calls a zig-zig motion (as the earth moves). Later he gives a different diagram (under subsection "Transformation of time") in relation to a "light clock". The light leaves a flashtube, hits a mirror and bounces back to a photocell. With the transverse motion of the light clock, it is assumed the light takes a diagonal path here too.

I think I have found the illustrations that you are referring to, but I do not have the abridged version of the lectures. I didn't even know there were any. That seems to be lecture I-15, and the figures are 15-2 Schematic diagram of the Michelson-Morley experiment, and 15-3 (a) A "light clock" at rest in the S' system, etc.

The same is true for them though--the light that does go out purely perpendicular...just misses, and doesn't participate in the experiment.



On 2002-04-27 12:53, Michele wrote:
But in his explanation of the M/M experiment, my impression, perhaps erroneous, was that the experiment didn't prove the lack of an ether per se, but merely the lack of an ether wind. I don't know if that's correct but it's just my reaction. I guess there could still be an ether which only acts as a light propagation medium and has no "wind" relative to any matter? Is that what the modern ether proponents are saying, I wonder?

Depends on the proponent, I'm sure, but the trouble starts as soon as they try to make some sort of inference based upon their proposition--the results might not differ hardly at all from the usual explanation. Why bother, then? It's like saying that on my back there is an invisible monkey, massless, quiet, and which doesn't do anything.

Karl
2002-Apr-29, 04:32 PM
On 2002-04-29 11:55, GrapesOfWrath wrote:
It's like saying that on my back there is an invisible monkey, massless, quiet, and which doesn't do anything.


You have one too?

Chip
2002-Apr-29, 05:07 PM
On 2002-04-27 02:06, Michele wrote:
"...if we consider a luminous object moving through space perpendicular to our line of sight (on earth), isn't it true that light that leaves the object and reaches earth (our eyes) will give the image of the light at the location from which it was emitted, instead of being translated by a distance equaling the time it takes for the light to reach our eyes multiplied by the object's lateral speed? What I mean to say is, I thought that the object would not give the emitted light any lateral momentum as a result of the object's transverse motion. Otherwise, when it arrived, the light would indeed indicate the object's true present position instead of its position of, say, 10 minutes ago if the object is 10 light minutes away from us.

Hi Michele,

Good questions. (Isn't this fun?) Remember that the failure of MM showed that the speed of light is constant regardless of the motion of either the light source or the observer. A simple, but highly deceptive concept with deep implications. There is also the whole matter of setting up coordinate systems between the observer and the observed light source relative to linear measured time.

Keeping in mind that the purpose of the experiment was to measure the effect of the "ether" relative to the direction of the Earth's motion, if Michelson-Morley had done their experiment under water, would not the refractive index of the water, which slows the speed of light, act as an artificial "ether"? /phpBB/images/smiles/icon_wink.gif

SiriMurthy
2002-Apr-29, 05:07 PM
On 2002-04-29 11:49, SeanF wrote:
The simple idea is that of inertia. The photon has the emitter's lateral motion before it's emitted, and inertia causes it to maintain that lateral motion.


Photons are basically massless. Where does the inertia come from?

SeanF
2002-Apr-29, 07:01 PM
On 2002-04-29 13:07, SiriMurthy wrote:


On 2002-04-29 11:49, SeanF wrote:
The simple idea is that of inertia. The photon has the emitter's lateral motion before it's emitted, and inertia causes it to maintain that lateral motion.


Photons are basically massless. Where does the inertia come from?


Photons have no rest mass, but since they're moving (at the speed of light, even!) they do have kinetic energy, which is the "same thing", ain't it?

Michele
2002-Apr-30, 02:12 AM
[quote]
On 2002-04-29 11:49, SeanF wrote:

The simple idea is that of inertia. The photon has the emitter's lateral motion before it's emitted, and inertia causes it to maintain that lateral motion.

It's just like with a gun. If the gun and target are both moving laterally at the same velocity, the bullet will still hit the target - it will continue to move laterally along with the gun and target while it is moving between the gun and the target.


Thank you, that pretty much answers my question, or I should say corrects my understanding. I was thinking, like SirMirthy, that light wouldn't have inertia in the way that mass does. But if light possess the inertia of the body from which it is emitted, then the diagrams I referred to do make sense. Incidentally, since light leaving distant stars, for example, possesses the momentum of those stars, do astronomers regularly incorporate the inertia of the light emitted from them in calculating their positions? Also, how is it known what the momentum vector is for distant "objects" by which I guess I mean galaxies, that are apparently motionless? So many questions, only one brain to think with.... /phpBB/images/smiles/icon_smile.gif

Michele
2002-Apr-30, 02:26 AM
[quote]
On 2002-04-29 11:55, GrapesOfWrath wrote:
I've given this some thought, as I'm not absolutely sure of what your question is. Are you aware of the phenomenon known as the aberration of light (http://www.harcourt.com/dictionary/def/1/7/3/0/17300.html)? It seems like that may be relevant. Stars emit light in all directions, but the photons that reach us do not arrive from the purely perpendicular direction.


If aberration is only the apparent displacement due to our (earth's) movement, then that's not what I'm referring to. I can understand that astronomers have to fit the earth's movement into equations to account for where the earth actually intersects the light. What I didn't know was whether the light leaving the object actually would have inertia in any direction other than its propagation as a result of the movement of its source.


I think I have found the illustrations that you are referring to, but I do not have the abridged version of the lectures. I didn't even know there were any. That seems to be


Yes, two volumes, "Six Easy Pieces" and "Six Not-So-Easy Pieces". Obvioulsy Relativity is in the latter. I also have tapes of most of the lectures but without diagrams some are hard to follow....

and the figures are 15-2 Schematic diagram of the Michelson-Morley experiment, and 15-3 (a) A "light clock" at rest in the S' system, etc. The same is true for them though--the light that does go out purely perpendicular...just misses, and doesn't participate in the experiment.


I'm not sure but I think the thought experiment for the light clock assumes a parallel beam, so all the light should be perpendicular. I've also seen other diagrams on the net where the same experiment is described using a single photon for illustration. (PHoton emitted, reflected, detected), so if it misses there is no detection.



Why bother, then? It's like saying that on my back there is an invisible monkey, massless, quiet, and which doesn't do anything.

Valid point, of course. But I think in this case, if there really were an invisible, unmeasurable ether it would still be doing something, i.e. acting as the medium for light wave propagation. That is, light would be definitely associated with a medium instead of empty space. But then I guess "empty" space would in fact be the ether and so your point stands.

Michele
2002-Apr-30, 02:33 AM
Sorry for the repost but I think the previous lack of formatting was confusing. Is there some easy way to draw those lines to delineate quoted material?

[quote]
On 2002-04-29 22:26, Michele wrote:
[quote]
On 2002-04-29 11:55, GrapesOfWrath wrote:
I've given this some thought, as I'm not absolutely sure of what your question is. Are you aware of the phenomenon known as the aberration of light (http://www.harcourt.com/dictionary/def/1/7/3/0/17300.html)? It seems like that may be relevant. Stars emit light in all directions, but the photons that reach us do not arrive from the purely perpendicular direction.


If aberration is only the apparent displacement due to our (earth's) movement, then that's not what I'm referring to. I can understand that astronomers have to fit the earth's movement into equations to account for where the earth actually intersects the light. What I didn't know was whether the light leaving the object actually would have inertia in any direction other than its propagation as a result of the movement of its source.


I think I have found the illustrations that you are referring to, but I do not have the abridged version of the lectures. I didn't even know there were any. That seems to be


Yes, two volumes, "Six Easy Pieces" and "Six Not-So-Easy Pieces". Obvioulsy Relativity is in the latter. I also have tapes of most of the lectures but without diagrams some are hard to follow....

and the figures are 15-2 Schematic diagram of the Michelson-Morley experiment, and 15-3 (a) A "light clock" at rest in the S' system, etc. The same is true for them though--the light that does go out purely perpendicular...just misses, and doesn't participate in the experiment.


I'm not sure but I think the thought experiment for the light clock assumes a parallel beam, so all the light should be perpendicular. I've also seen other diagrams on the net where the same experiment is described using a single photon for illustration. (PHoton emitted, reflected, detected), so if it misses there is no detection.



Why bother, then? It's like saying that on my back there is an invisible monkey, massless, quiet, and which doesn't do anything.

Valid point, of course. But I think in this case, if there really were an invisible, unmeasurable ether it would still be doing something, i.e. acting as the medium for light wave propagation. That is, light would be definitely associated with a medium instead of empty space. But then I guess "empty" space would in fact be the ether and so your point stands.

James
2002-Apr-30, 02:39 AM
On 2002-04-29 22:33, Michele wrote:
Sorry for the repost but I think the previous lack of formatting was confusing. Is there some easy way to draw those lines to delineate quoted material?

Michele, just don't delete those [ / quote ] at the end of each quote.

Michele
2002-Apr-30, 02:46 AM
[quote]
On 2002-04-29 13:07, Chip wrote:
Good questions. (Isn't this fun?) Remember that the failure of MM showed that the speed of light is constant regardless of the motion of either the light source or the observer. A simple, but highly deceptive concept with deep implications. There is also the whole matter of setting up coordinate systems between the observer and the observed light source relative to linear measured time.


Just for discussion sake, would the M/M experiment theoretically also have given a null result if light *didn't* have any momentum from the moving earth (relative to an "ether")?

Michele
2002-Apr-30, 02:48 AM
On 2002-04-29 22:39, James wrote:

Michele, just don't delete those [ / quote ] at the end of each quote.



Oh, I see, thanks!

Michele
2002-Apr-30, 03:36 AM
On 2002-04-29 15:01, SeanF wrote:
Photons have no rest mass, but since they're moving (at the speed of light, even!) they do have kinetic energy, which is the "same thing", ain't it?


When a photon reflects off an object, is its reflection perfectly "elastic"? In other words, it is traveling at speed c an instant before hitting whatever reflects it, and at *some* point (the actual theoretical point where it changes direction), its velocity would be zero, and if it were, it would have to accelerate up to c again. Well, I realize that what I'm describing is the way matter works, so maybe it doesn't apply to photons. So, is it that such a point (of changing direction where the velocity is zero) doesn't really exist for electromagnetic energy? Maybe this is one aspect where the particle description of light doesn't work? I think I can visualize how a wave could have constant speed one instant before reflection and one instant after (without decelerating and accelerating), but not how a [b]particle[b/
] could.

SimonCB
2002-Apr-30, 10:04 AM
Your right that in the case of reflection its best not to think about the photon as a particle.

In the case of reflection a photon is not bouncing of a solid object. It is being absorbed and re-emmitted. So it is a real sense not the same photon coming back the other way. This is explanation of both colour and spectral absorbtion features.

Simon

GrapesOfWrath
2002-Apr-30, 12:32 PM
On 2002-04-29 22:33, Michele wrote:
What I didn't know was whether the light leaving the object actually would have inertia in any direction other than its propagation as a result of the movement of its source.

And what do you think would be the consequence of that? Its direction of propogation is its direction of propogation, simply, whether that is the direction you think it should be or not.



I think I have found the illustrations that you are referring to, but I do not have the abridged version of the lectures. I didn't even know there were any. That seems to be

Yes, two volumes, "Six Easy Pieces" and "Six Not-So-Easy Pieces". Obvioulsy Relativity is in the latter.

Ah, I thought about those. I don't think that they are abridged, though. I think they are just excerpted. Six Easy Pieces is just chapters I-1, I-2, I-3, I-4, I-7, and I-37 of the first volume of Feynman's red books.


I'm not sure but I think the thought experiment for the light clock assumes a parallel beam, so all the light should be perpendicular. I've also seen other diagrams on the net where the same experiment is described using a single photon for illustration. (PHoton emitted, reflected, detected), so if it misses there is no detection.

What do you mean by an assumption of a parallel beam? Even in the single photon setup, the photon is assumed to strike the reflecter, so no misses involved in the actual thought experiment.



Why bother, then? It's like saying that on my back there is an invisible monkey, massless, quiet, and which doesn't do anything.

Valid point, of course. But I think in this case, if there really were an invisible, unmeasurable ether it would still be doing something, i.e. acting as the medium for light wave propagation. That is, light would be definitely associated with a medium instead of empty space. But then I guess "empty" space would in fact be the ether and so your point stands.

Exactly. Empty space means "ether" to some, not to others. I think that there is a quote from Einstein in his later years, to that effect. However, if we can find no additional effect of that empty space, then "empty space" it is.

Russ
2002-Apr-30, 01:45 PM
Hail & welcome aboard Michele. I'm no expert but would like to dip my oar in the water so to speak.

What I think is getting lost in this discussion is that light behaves as both particle and wave. This property makes talking about light as only a particle..uuumm...dangerous. I'll use an example from my physics class back in the early 70's. I hope I can present it as clearly as my prof. did.

Imagine a big pond of water. All of the little H2O molecules represent photons. They can be move individually or as part of a wave. A boat traverses in front of you standing on the beach. The wake (waves) from the boat impinge on the shore as both individual particles and as a whole wave. So a selected individual particle of water (light) may "miss" you but you still see the wave hit the shore where you stand. All of the other effects (doplar, absorbtion, emission, etc.) of wave motion still work and apply.

The analogy of a bullit to a photon is apt in some ways but pretty loose at best. In the case of a reflection, the bullit (photon) does not stop and change direction accelerating back to the speed of light. It simply "skips" off the reflector in another direction, losing some energy (heat, motion, etc.) but traveling no slower at any time. In a refraction (bullit punches through target & is deflected) system, the bullit may slow as it passes through the target medium but stops losing energy as soon as it emerges from the target.

I hope I have not muddied the water for you. Again, welcome aboard the board. /phpBB/images/smiles/icon_biggrin.gif

Michele
2002-Apr-30, 03:04 PM
Wow, after writing a long response I inadvertently put in the wrong password and lost everything. Bugger. This version is going to be shorter.... /phpBB/images/smiles/icon_frown.gif



And what do you think would be the consequence of that? Its direction of propogation is its direction of propogation, simply, whether that is the direction you think it should be or not.


I'm assuming light emitted outward has a fixed direction of propagation, whether or not it has lateral motion.




Ah, I thought about those. I don't think that they are abridged, though. I think they are just excerpted. Six Easy Pieces is just chapters I-1, I-2, I-3, I-4, I-7, and I-37 of the first volume of Feynman's red books.

Yeah, excerpts. I meant the collections are abridged (not complete).1



What do you mean by an assumption of a parallel beam? Even in the single photon setup, the photon is assumed to strike the reflecter, so no misses involved in the actual thought experiment.


I mean that the only light being considered is the light that leaves perpendicular to the lateral direction of movement, which would be all parallel waves, right? So if the outgoing perpendicular light you mentioned would miss the reflector, isn't that the only light that's being considered?

GrapesOfWrath
2002-Apr-30, 03:17 PM
On 2002-04-30 11:04, Michele wrote:
Wow, after writing a long response I inadvertently put in the wrong password and lost everything. Bugger.

If you use the back button, your response will still be there. At least with my browser, YMMV.




And what do you think would be the consequence of that? Its direction of propogation is its direction of propogation, simply, whether that is the direction you think it should be or not.


I'm assuming light emitted outward has a fixed direction of propagation, whether or not it has lateral motion.

Certainly, but then the answer to your implied question of "What I didn't know was whether the light leaving the object actually would have inertia in any direction other than its propagation as a result of the movement of its source," is no, it does not. Nothing in addition to its direction of propagation.




What do you mean by an assumption of a parallel beam? Even in the single photon setup, the photon is assumed to strike the reflecter, so no misses involved in the actual thought experiment.


I mean that the only light being considered is the light that leaves perpendicular to the lateral direction of movement, which would be all parallel waves, right?

Of course not, in the frame of reference of the lab. That's pretty obvious from the diagram. Perhaps that is the source of your confusion.


So if the outgoing perpendicular light you mentioned would miss the reflector, isn't that the only light that's being considered?

Definitely not.

I'm not sure why you're insisting on that. Light doesn't just emit perpendicular to an emitter's motion. OTOH, you can see how, in the reference frame of the moving emitter, that light does stay perpendicular, no?

Wiley
2002-Apr-30, 07:36 PM
Welcome, Michele.

I wanted to add a few things to what Grapes and the others have said which will simplify things (I hope). First this is a "classical" problem, so we don't have to treat light as a particle, we can treat it as a wave. We should get the same answer regardless of what light "is", so think of light in the manner most comfortable to you.

Another trick is to interchange the source and the observer. In your example, the source was moving, and the observer still. Let the source be still and the observer moving. The essence of relativity is that these are equivalent. Personally, I find it easier to visualize the field from a stationary source. Again, whatever is easier for you.

Now to the more fun stuff:


Otherwise, when it arrived, the light would indeed indicate the object's true present position instead of its position of, say, 10 minutes ago if the object is 10 light minutes away from us.

The field does indeed point to the source's true position. However this does not violate SR. Imagine that source stopped suddenly, it would take 10 minutes for the observer to know the source had stopped. The field at the observer 9:59 minutes after the source stopped points to where the source would be if had not stopped. (This is part of thread "Flat Earth Society ..., best to stay away from that one. /phpBB/images/smiles/icon_smile.gif)

Or, as SeanF suggested, think of light as a particle that has lateral momemtum.

Hope this helps,

Michele
2002-May-01, 07:20 AM
On 2002-04-30 11:17, GrapesOfWrath wrote:


I'm assuming light emitted outward has a fixed direction of propagation, whether or not it has lateral motion.

Certainly, but then the answer to your implied question of "What I didn't know was whether the light leaving the object actually would have inertia in any direction other than its propagation as a result of the movement of its source," is no, it does not. Nothing in addition to its direction of propagation.


But, I have just learned (if I understand it correctly), that the light is given lateral momentum by its moving environment, a direction which is a different direction than its propagation, no? I must be missing something crucial because I see two separate vectors, the outward vector of the light which was propagated directly outward, and the transverse vector which is entirely due to the movement of the environment. Even if the light moves diagonally overall as a result, isn't its direction of propagation still outward (perpendicular)? (Like, if you shoot a bullet outward in the same moving environment, its overall path is diagonal but the bullet is always pointing perpendicularly outward. Light is directional like that, isn't it?



Michele:
I mean that the only light being considered is the light that leaves perpendicular to the lateral direction of movement, which would be all parallel waves, right?

Gr of W:
Of course not, in the frame of reference of the lab. That's pretty obvious from the diagram. Perhaps that is the source of your confusion.
[/quote]

I'll admit to being confused. Here's where I'm at: I see the frame of reference of the lab (MM experiment?) as being that of the system itself (since both on the earth moving through the ever-elusive "ether"). The light source sends out photons in every direction imaginable, most of which are not considered in the experiment as you pointed out. To me, that means that all the photons that just happened to be moving parallel to each other in the "perpendicular" direction (after the half mirror) are what is being considered. For the light-clock diagram, the same thing only from the viewpoint of the rider -- all the parallel propagating photons within the width of the reflector hit it and come back, while the others are going in different directions and don't participate. Someone not on the vehicle would "see" the beam moving diagonally, but I am finding it difficult to grasp that the *direction of propagation* would be diagonal even for him. That only because of the directional nature of light waves. This may be weird, but if the outside observer wanted to polarize the light beam, he would know that he would have to (if he could somehow do it instantaneously as the vehicle passed by him) place the polarizing plate oriented normal to what the rider considers the perpendicular direction. If he oriented it normal to the diagonal which he sees as the "direction of propagation", the light wouldn't polarize properly, isn't that right?




I'm not sure why you're insisting on that. Light doesn't just emit perpendicular to an emitter's motion. OTOH, you can see how, in the reference frame of the moving emitter, that light does stay perpendicular, no?


Oh you must have meant light perpendicular from the outside observer's reference misses the reflector. WHich I think would be diagonally backwards for the rider. You see, what I thought before was that even light perpendicular from the rider's reference frame ought to miss the reflector (if it lacked the lateral momentum).

Actually it just struck me that even if, just for the sake of argument, it did lack lateral momentum, there would still be a beam at an angle, which is probably the arccosine (?) of the lateral speed /c, that would still hit the reflector and come back. I wonder how could we distinguish that situation from our present understanding that light is given momentum by its emitter. I mean, how could we know whether the light beam that was detected was the component emitted perpendicular (with lateral momentum) and not a component directed out at a forward angle (with no lateral momentum)? This is all speaking in the frame reference of the rider moving with the light clock...

GrapesOfWrath
2002-May-01, 07:29 AM
On 2002-05-01 03:20, Michele wrote:
I wonder how could we distinguish that situation from our present understanding that light is given momentum by its emitter. I mean, how could we know whether the light beam that was detected was the component emitted perpendicular (with lateral momentum) and not a component directed out at a forward angle (with no lateral momentum)?

It's all the same thing, as near as I can tell. The direction of propagation, whether you break it into components, or not.

Take a "light beam" and then shift your frame of reference by 45 degrees. It then has two equal components, at right angles to each other.

Michele
2002-May-01, 07:40 AM
On 2002-04-30 15:36, Wiley wrote:


Otherwise, when it arrived, the light would indeed indicate the object's true present position instead of its position of, say, 10 minutes ago if the object is 10 light minutes away from us.

The field does indeed point to the source's true position. However this does not violate SR. Imagine that source stopped suddenly, it would take 10 minutes for the observer to know the source had stopped. The field at the observer 9:59 minutes after the source stopped points to where the source would be if had not stopped. (This is part of thread "Flat Earth Society ..., best to stay away from that one. /phpBB/images/smiles/icon_smile.gif)


Thanks, that's interesting. To think of all the things astronomers have to take into account just to get an object's position is rather mind-boggling. It seems to me also that at 9:59 minutes after it stops, since the light shows it to be where it would have been if it had kept going, then between 9:59 and 10:00 it's image would have to move backward to it's actual present position (where it is now stationary), at a rate corresponding to the object's deceleration to stop. Is that what actually happens?

Michele
2002-May-01, 09:52 AM
On 2002-05-01 03:29, GrapesOfWrath wrote:


On 2002-05-01 03:20, Michele wrote:
I wonder how could we distinguish that situation from our present understanding that light is given momentum by its emitter. I mean, how could we know whether the light beam that was detected was the component emitted perpendicular (with lateral momentum) and not a component directed out at a forward angle (with no lateral momentum)?

It's all the same thing, as near as I can tell. The direction of propagation, whether you break it into components, or not.



Only that SeanF suggested that Relativity would be in trouble if photons don't carry the inertia of their emitters, and so there ought to be a way to confirm that it's really the (inside frame) perpendicular light with inertia and not forward angular light without inertia. Using my own logic against this, the direction of the light propagation ought to do that, but if the direction of propagation is simply the direction its moving, how can we know which situation is correct?

GrapesOfWrath
2002-May-01, 01:25 PM
On 2002-05-01 05:52, Michele wrote:
Using my own logic against this, the direction of the light propagation ought to do that, but if the direction of propagation is simply the direction its moving, how can we know which situation is correct?

That is where I lose the logic about the point being made. "The direction of propagation = the direction its moving" seems like a tautology to me. What other interpretation would there be?

2002-May-01, 01:53 PM
<a name="20020501.5:46"> page 20020501.5:46 aka two scents
On 2002-05-01 09:25, GrapesOfWrath wrote: To: 2 cents
particle : WAVE in the water example
its invisioned that the Waves composed of the particles.
In lights case it may well be that its
Just the oposite CASE
the particles may be composed of Wave
and thus it {may} depend upon the detector
a cork may respond to the wave
while a big poat only to the particles?

SeanF
2002-May-01, 03:01 PM
On 2002-04-30 15:36, Wiley wrote:

The field does indeed point to the source's true position. However this does not violate SR. Imagine that source stopped suddenly, it would take 10 minutes for the observer to know the source had stopped. The field at the observer 9:59 minutes after the source stopped points to where the source would be if had not stopped.



Wiley, you've confused me (what do you mean, "Again?"). /phpBB/images/smiles/icon_smile.gif Can you explain what you mean by the "field" "pointing to" a "position"?

It seems like you're suggesting the light would appear to be coming from a location which the emitter never occupies, and I'm not sure I follow that . . . /phpBB/images/smiles/icon_frown.gif


_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2002-05-01 11:03 ]</font>

SeanF
2002-May-01, 03:15 PM
On 2002-05-01 09:25, GrapesOfWrath wrote:


On 2002-05-01 05:52, Michele wrote:
Using my own logic against this, the direction of the light propagation ought to do that, but if the direction of propagation is simply the direction its moving, how can we know which situation is correct?

That is where I lose the logic about the point being made. "The direction of propagation = the direction its moving" seems like a tautology to me. What other interpretation would there be?


GoW, I think I might've picked up something from Michele's mention of a polarizing filter above (Michele, please correct anything I say here that doesn't match with what you're thinking).

Take a piece of paper and make three points A, B, and C that form a right triangle with B being at the right angle. If your emitter is at A and your receiver is at B, a polarizing filter would be placed on the line AB perpendicular to the line AB. If the receiver is at C, the polarizing filter would be placed on line AC perpendicular to the line AC.

Let's consider our experiment with the emitter and receiver on the train car. To an observer on the car, there is no lateral motion - the emitter is at A and the receiver is at B. He places the filter between them, perpendicular to AB.

An observer off the train car sees the lateral motion. He sees the emitter at A when the light is emitted and the receiver at C when the light is received. The lateral motion would also carry the filter so that it would indeed be on the line AC at the correct time, but it would not be perpendicular to AC. Therefore, the two observers should disagree on whether or not the light comes through the filter correctly polarized.

Is that what you're thinking, Michele?

GrapesOfWrath
2002-May-01, 03:39 PM
Ah, but remember, line AC and line AB, are the same line, no? Just different reference frames.

Wiley
2002-May-01, 10:23 PM
On 2002-05-01 11:01, SeanF wrote:
Wiley, you've confused me (what do you mean, "Again?"). /phpBB/images/smiles/icon_smile.gif Can you explain what you mean by the "field" "pointing to" a "position"?

It seems like you're suggesting the light would appear to be coming from a location which the emitter never occupies, and I'm not sure I follow that . . . /phpBB/images/smiles/icon_frown.gif


SeanF, just trying to keep you on your toes. /phpBB/images/smiles/icon_smile.gif The direction of the electric field gives the true position of the source, not the retarded position, provided both source and observer are in inertial frames. When the source starts accelerating or decelerating, the electric field will no longer give the true position.

Consider a stationary charge (v = 0, a = 0) at the origin. The electric field is radially directed, E = (constants) r/R<sup>2</sup> where r is the unit vector in radial direction. The field points "away" from the charge. Er...of course, in the example in my previous post, the charge is negative. /phpBB/images/smiles/icon_smile.gif

An observer moving with constant velocity, i.e. in a valid inertial frame, would see field pointing directly away from the source. We could now invoke the principle of relativity and claim that for a stationary observer and a source moving at a constant velocity, the field points away from the charge is, not was. However, this would be taking the easy way out.

The velocity field for a moving charge is

E = (constants) (1 - v^2) R<sup>-2</sup>(1 - r*v)<sup>-3</sup> (r - v)

where R is the distance from the source to the observer at the retarded time and r is the unit vector from the source to the observer at the retarded time and v is the normalized velocity of the source. Retarded time is the time when the light reaching the observer was actually emitted by the source.

The important part for this discussion is the direction of the field, r - v. Obviously the field does not point towards the position of the charge when the light was emitted, that direction is r. Does it point to the instantaneous position of the charge? A quick sketch of the vector -(r - v) will show it points in that general direction.

Let's simplify things a bit. Assume the observer is at (0,y,0) and the source is moving along the x axis. To make the algebra a little easier, we'll find the direction of the observer's field when the source reaches the origin (0,0,0).

The difference between retarded time, t<sub>r</sub>, and observer time, t, is R/c. Thus the position of source at the retarded time is (-vc(t - t<sub>r</sub>,0,0) = (-vR,0,0). (Recall v is normalized, hence we need the c.) The unit vector r is thus r = vx + (y/R)y. The direction of the E field is r - v = (y/R)y, the field points directly away from the true, not the retarded, position.



<font size=-1>[ This Message was edited by: Wiley on 2002-05-01 18:38 ]</font>

Wiley
2002-May-01, 10:40 PM
On 2002-05-01 03:40, Michele wrote:
It seems to me also that at 9:59 minutes after it stops, since the light shows it to be where it would have been if it had kept going, then between 9:59 and 10:00 it's image would have to move backward to it's actual present position (where it is now stationary), at a rate corresponding to the object's deceleration to stop. Is that what actually happens?



Exactly.

Michele
2002-May-02, 01:21 AM
On 2002-05-01 09:25, GrapesOfWrath wrote:


On 2002-05-01 05:52, Michele wrote:
Using my own logic against this, the direction of the light propagation ought to do that, but if the direction of propagation is simply the direction its moving, how can we know which situation is correct?

That is where I lose the logic about the point being made. "The direction of propagation = the direction its moving" seems like a tautology to me. What other interpretation would there be?


It seems to me that the direction the light was emitted in (the direction the emitter was actually pointed) could not change simply as a result of moving or not moving. On the other hand, if there is no discernible difference between the perpendicular beam and the diagonal beam in terms of *direction*, then won't that put us in the quandry I described earlier, i.e. we can't tell whether light is given momentum by its emitter or not? All the experimenter knows is he emitted a beam and light was detected at his detector -- he has no way of knowing what sort of phenomenon of motion was actually involved, so unless there is an orientation of the waves themselves, it could have been light emitted perpendiculary outward and translated laterally by the movement, or it could have been light directed at a specific angle that was completely unaffected by the movement of the emitter. Or so it seems to my early morning brain cells.

<font size=-1>[ This Message was edited by: Michele on 2002-05-01 21:37 ]</font>

Michele
2002-May-02, 01:29 AM
On 2002-05-01 09:53, HUb' wrote:
<a name="20020501.5:46"> page 20020501.5:46 aka two scents
On 2002-05-01 09:25, GrapesOfWrath wrote: To: 2 cents
particle : WAVE in the water example
its invisioned that the Waves composed of the particles.
In lights case it may well be that its
Just the oposite CASE
the particles may be composed of Wave
and thus it {may} depend upon the detector
a cork may respond to the wave
while a big poat only to the particles?


When hearing my first significant words on light phenomena in first year college, I remember my reaction was "what's all the fuss about understanding dual nature of light -- why, water is composed of particles yet acts in waves, so does air..." Then some years later I heard a comment by some famous physicist that if you think you understand the dual nature, you don't. And so I definitely don't.... /phpBB/images/smiles/icon_smile.gif

Michele
2002-May-02, 01:35 AM
On 2002-05-01 11:15, SeanF wrote:
An observer off the train car sees the lateral motion. He sees the emitter at A when the light is emitted and the receiver at C when the light is received. The lateral motion would also carry the filter so that it would indeed be on the line AC at the correct time, but it would not be perpendicular to AC. Therefore, the two observers should disagree on whether or not the light comes through the filter correctly polarized.

Is that what you're thinking, Michele?


Yes, that's the gist of it. I know a matter analogy isn't always valid but if it were a knife being propelled outward, both viewpoints would have to at least agree on the most lethal direction (the direction of "propagation").

Michele
2002-May-02, 02:37 AM
Incidentally, another part of the light clock example I don't get is the following (This might be flame material, since I ought to realize its only a theoretical example). But still, I think that when we talk about an outside stationary observer and what he sees as the light clock whizzes by, it would all depend on where the observer is.

For example, the observer has two eyes obviously but we assume one point. From any given point distant from the path of the light clock, he would indeed see a zig-zag pattern of movement, but the isoceles triangle defined by this zig-zig and the path of movement as the base, if he is to judge by how the light actually appears to move, would have a different appearance depending on how distant he is from the path, and so the time dilation equation would be different, no? For an extreme case where the observer is standing right next to the path of movement, for him the light path would be essentially what the rider observes, i.e. straight up and down (perpendicular) movement (of course he'd have to "turn his head" 180 degrees once it passed him). So no time discrepancy should arise there since the light traveled the same *apparent* distance in the same time. I'm only talking about how the light *appears* to move to the observer because of his angle of observation. This might be irrelevant, but isn't the original thought experiment based on how the light *appears* to move straight up and down for the rider whereas the light is *really* (as fas as the rest of the universe is concerned) not taking such a simple path. ?

GrapesOfWrath
2002-May-02, 08:13 AM
On 2002-05-01 21:29, Michele wrote:
When hearing my first significant words on light phenomena in first year college, I remember my reaction was "what's all the fuss about understanding dual nature of light -- why, water is composed of particles yet acts in waves, so does air..."

And, of course, each particle acts in a wavelike manner, by itself.

SeanF
2002-May-02, 01:50 PM
On 2002-05-01 22:37, Michele wrote:
Incidentally, another part of the light clock example I don't get is the following (This might be flame material, since I ought to realize its only a theoretical example). But still, I think that when we talk about an outside stationary observer and what he sees as the light clock whizzes by, it would all depend on where the observer is.

For example, the observer has two eyes obviously but we assume one point. From any given point distant from the path of the light clock, he would indeed see a zig-zag pattern of movement, but the isoceles triangle defined by this zig-zig and the path of movement as the base, if he is to judge by how the light actually appears to move, would have a different appearance depending on how distant he is from the path, and so the time dilation equation would be different, no? For an extreme case where the observer is standing right next to the path of movement, for him the light path would be essentially what the rider observes, i.e. straight up and down (perpendicular) movement (of course he'd have to "turn his head" 180 degrees once it passed him). So no time discrepancy should arise there since the light traveled the same *apparent* distance in the same time. I'm only talking about how the light *appears* to move to the observer because of his angle of observation. This might be irrelevant, but isn't the original thought experiment based on how the light *appears* to move straight up and down for the rider whereas the light is *really* (as fas as the rest of the universe is concerned) not taking such a simple path. ?



Michele, I think you're misinterpreting the concept of how the light "appears" to move to the two observers.

Consider the situation where the light is being bounced off a mirror and returning to the same point (your emitter and receiver are located together).

If the whole shebang is not moving laterally, then the light simply moves straight "up" to the mirror and straight "down" back to the receiver, right?

If, however, the whole thing is moving laterally, then in order for the light to get back to the receiver the light must also be moving laterally, i.e. at an angle.

The angle would be dependent on the speed of the lateral motion and nothing else - not the position of the observer who sees the whole thing moving. If the mechanism moves laterally 500 ft in the amount of time it takes the light to make it's trip, then the light must have moved laterally 500 ft.

When talking about Relativity and talking about what two different observers "see," it's not really a matter of how anything physically hits their eyes. It's just a matter of how the events that occur would be logically considered to have occured.

I would never survive as a teacher - I hope I'm making some kind of sense here. /phpBB/images/smiles/icon_smile.gif

Michele
2002-May-02, 03:09 PM
On 2002-05-02 09:50, SeanF wrote:

The angle would be dependent on the speed of the lateral motion and nothing else - not the position of the observer who sees the whole thing moving. If the mechanism moves laterally 500 ft in the amount of time it takes the light to make it's trip, then the light must have moved laterally 500 ft.

When talking about Relativity and talking about what two different observers "see," it's not really a matter of how anything physically hits their eyes. It's just a matter of how the events that occur would be logically considered to have occured.

I would never survive as a teacher - I hope I'm making some kind of sense here. /phpBB/images/smiles/icon_smile.gif



Yes, you are actually. Ok, the observer = logical occurrence of events. I think I got it.

GrapesOfWrath
2002-May-02, 03:15 PM
<font size=-1>[Fixed equation, ah, heck]</font>

<font size=-1>[ This Message was edited by: GrapesOfWrath on 2002-05-02 22:56 ]</font>

Wiley
2002-May-02, 10:48 PM
On 2002-05-02 09:50, SeanF wrote:
The angle would be dependent on the speed of the lateral motion and nothing else - not the position of the observer who sees the whole thing moving. If the mechanism moves laterally 500 ft in the amount of time it takes the light to make it's trip, then the light must have moved laterally 500 ft.

When talking about Relativity and talking about what two different observers "see," it's not really a matter of how anything physically hits their eyes. It's just a matter of how the events that occur would be logically considered to have occured.



Observers in the same inertial frame should "see" the same triangle created by the zigzagging. Hopefully both of the observer's eyes are in the same inertial frame. If not, Ouch! Serious headache. /phpBB/images/smiles/icon_smile.gif

<font size=-1>[ This Message was edited by: Wiley on 2002-05-02 18:59 ]</font>

Wiley
2002-May-02, 11:11 PM
I wanted to add a little bit on polarization. There are really two types of polarizations, horizontal and vertical. There are others polarizations, like circular, but they are linear combinations of horizontal and vertical.

Consider the case of moving source and a stationary observer, both in the x-y plane. Recall that light is a transverse wave, which means the electric and magnetic fields point 90 degrees from the direction of the wave. A horizontally polarized wave would have an electric field oscillating in the x-y plane, and a vertically polarized wave would have an electric field oscillating in the z direction. When you increase or decrease the speed of the source, you will change the phase of the wave, i.e. the direction of the field in the plane, but it will still oscillate in the same plane. (side note: phase change -> Doppler effect.) Does this make sense?

Polarizing filters would not detect any difference between a stationary source and a source moving with a constant velocity.

<font size=-1>[ This Message was edited by: Wiley on 2002-05-02 19:12 ]</font>

Ring
2002-May-05, 03:42 AM
Just a few minor points:

1. A single photon does not have inertia but it does have momentum.

2. The electric field, to first order, has a retardation correction. If this weren't so then there could be no such thing as an electromagnetic wave.

3. The electromagnetic wave (light, a photon) does not have a correction and most definitely is retarded.

Ring
2002-May-05, 03:57 AM
Actually any number of photons do not have inertia if there is no center of momentum frame.

kjavds
2002-May-06, 12:51 AM
Micelle: I happen to know PRECISELY what you are talking about, and the answer is a resounding NO! light does NOT have 'lateral momentum'. The reason that those diagrams show a slanted line for the light path as seen from an alien observatory is simply a skewing that needs be depicted for the new viewpoint. After all, the light clearly started where the flashlight was, then moved to the mirror, then returned to where the flashlight now is, and that makes a triangle shape, plain and simple. You are 100% correct that light CANNOT have 'lateral momentum' because all light paths are independnt of the motion of their source. But from a bystander's viewpoint, this fundamental accomodation must be made in order to rightly diagram the scenario. I do not approve either of their methodology for teaching and diagramming Relativity. I HIGHLY recommend the brief treatise at http://www.ezrelativity.tk for complete clarity on the subject.

2002-May-07, 10:48 AM
<a name="20020507.2:41"> page 20020507.2:41 aka "Amplydine"{sp}
On 2002-05-04 23:42, Ring wrote: To: 10 CAUAC 13 UO
Just a few minor points:

1. A single photon does not have inertia but it does have momentum.
HUb' don't know & don't much care {See June}
2. The electric field, to first order, has a retardation correction. If this weren't so then there could be no such thing as an electromagnetic wave.
HUb' sats "Hmm? {si' July}
3. The electromagnetic wave (light, a photon) does not have a correction and most definitely is retarded.
HUb' inplies that in an _-Amplydine-_ the "rotating" Magnetic field (induced by the shorted armature) MUST have some amount
of residual drag to its surrounding housing or it just would not be cut by the secondary {90deg} winding?