Creating a Target with 3 Photons in SR Propagating Sphere of Light

[Reff]

I assume in an SR example of a propagating sphere of light from a pinhead size event that its propagation symetry is independent of the frame which created it. The point on the frame which created the event must be somewhere within the sphere.

Use three photons on the sphere to create a target. 1. Any one photon 2. A photon heading out on the reciprocal of the first photon. 3. A photon heading out at 90 degrees to the line formed by the first two photons.

During propagation of the three photons take a small target of some mass and hold it on the reciprocal of the three photons for a short duration and release.
The target will now be inertial and remain on the reciprocal of the three photons.

Is it reasonable to suggest the photons move at c from the point marked by the target to the edge of the sphere but not at c from the point on the frame they left unless that point remains with the target.

Are these reasonable assumptions or at what specific point am I mistaken. I find this can be used as the basis for simple SR calculations.

 

[Dale]

Hi Reff, welcome to PF!

During propagation of the three photons take a small target of some mass and hold it on the reciprocal of the three photons for a short duration and release.
The target will now be inertial and remain on the reciprocal of the three photons.

You mean place the target at the origin?

Is it reasonable to suggest the photons move at c from the point marked by the target to the edge of the sphere but not at c from the point on the frame they left unless that point remains with the target.

I don't know what you are saying here.

 

[Reff]

Hi DaleSpam
Thanks for your welcome and thanks for having a go at understanding my description. It is difficult to try and follow another persons description.
Re placing a target at the origin.
If the origin was the tip of an object moving at .6c and we freeze the sphere and object I believe all the photon reciprocals generated from its event do not return to that origin. During propagation the ship has moved from the reciprocals. Placing a target at the origin for me means holding it at the reciprocal of all the photons within the sphere. Holding it there for some of the propagation means it will stay on the reciprocals and not move away because it was only momentarily at that position. At any stage during propagation the now inertial target will remain on the reciprocal of all the photons on the sphere and using just three is enough to demonstrate that point.
The second part re moving at c. Well. freeze the target sphere at say two meters radius and measure the distance from the target to any photon on the edge. they will all be two meters.
Do this for the moving frame origin and the point of origin on the frame will have moved during propagation so what I believe I am trying to say is, the photon sphere is the absolute datum for c and therefore a passenger on the moving frame would be incorrect in believing from the point of origin of the event on the tip of his frame to its propagating sphere has a perfect symetry like the target photon sphere. If you are still interested I can go a little further with some simple geometry to show c within a target sphere as opposed to c within a moving frame where time dilation must be the consequence, ie .6c frame has a .8t clock. I believe my view is not the norm and am open to correction.

 

[Drakkith]

Your view hasn't taken time dilation or length contraction into account, at least not correctly. From the frame of the moving observer, the photons all move at the same speed away and do indeed form a sphere. But they also form a sphere in the stationary observer. The stationary observer will also measure the photons at different frequencies depending on what direction they are moving in relation to the direction the source is moving.

 

[Dale]

A sphere of light in one frame is a sphere of light in all frames and the rays emanate from the origin in all frames. It's weird, but that is how it works.

 

[Reff]

Hi DaleSpam and Drakkith
Thanks for your replys.
Yes I am comfortable with time dilation as my geometry gives the correct answers and I feel I can see it clearly. With length contraction I am comfortable as I see SR is a function of the lateral part of a moving frame. For frequencies I believe that photons from an event have not had time to form a wave as events are of such a short duration.
You both mention the symetry of photon spheres. Re DaleSpam.

A sphere of light in one frame is a sphere of light in all frames and the rays emanate from the origin in all frames.

I hope you will read on If I disagree.
Lets look at two examples. Let's try the target frame first. After some propagation, photons on the sphere will have a very specific heading so they will have also have specific reciprocals. Reverse the photons and look at their intersections with the target. Throughout the whole exercise, of propagation and reversal the target mass is inertial. It is always on the reciprocals.
Now look at the moving inertial frame. Reverse the photon direction to form the photon intersection. In this case the moving frame must also instantly reverse direction to get back to the point of origin intersection. I think the observers in both frames would agree there is a big difference.
Both observers believe they know the event point origin on their frame but one observer will find his frame will need to reverse its direction in an instant to arrive back to where the photons really did start from. I believe photons move at c only from the crossing of their reciprocals which like headings are precise.

 

[Dale]

A sphere of light in one frame is a sphere of light in all frames and the rays emanate from the origin in all frames.

I hope you will read on If I disagree.

I did read on, but your disagreement is factually incorrect.

The equation of a spherical pulse of light emanating from the spatial origin at t=0 is given by:
c^2 t^2 = x^2 + y^2 + z^2

Applying the Lorentz transform ( http://en.wikipedia.org/wiki/Lorentz_transformation ) in the standard configuration we get:
c^2 \gamma ^2 \left(t&#039;-\frac{v x&#039;}{c^2}\right)^2=\gamma ^2 \left(x&#039;-v<br /> t&#039;\right)^2+\left(y&#039;\right)^2+\left(z&#039;\right)^2

Which simplifies to:
c^2 \left(t&#039;\right)^2=\left(x&#039;\right)^2+\left(y&#039;\right)^2+\left(z&#039;\right)^2

So a spherical pulse of light centered on the origin in one frame is a spherical pulse of light centered on the origin in any other frame. If you trace the rays back in any frame you find that any two non-colinear rays intersect at the origin of that frame. The target which is at rest on the intersection/origin for one frame is not at rest on the intersection/origin for any other frame.

If you still disagree then I challenge you to work the following problem:

1) In the frame where the source is at rest write the equation of motion for any two non-colinear rays of light.
2) Apply the Lorentz transformation on the equations to get the equations of motion for the rays in any other frame.
3) Find the intersection of the rays in the first frame.
4) Find the intersection of the rays in the second frame.

 

[ghwellsjr]

Reff, I've never heard of the reciprocal of a photon. Here's some possibilties based on your posts of what I think you might mean:

1) When a photon is emitted from a source, its reciprocal is another photon emitted from the same source at the same time but in the opposite direction.

2) When a photon is emitted from a source, its reciprocal is another photon emitted from the same source in the opposite direction but at a later time.

3) When a photon is emitted from a source and later strikes a target, such as a reflector, a second photon is emitted from the target in the opposite direction and travels back to the source of the original photon.

4) When a photon is emitted from a source and later strikes a target, a second photon is emitted from the target in some direction not necessarily related to the first photon, say for example, at 90 degrees.

5) Something else entirely.

Please let us know what you mean by the reciprocal of a photon and provide a link or a reference to its definition unless you made it up on your own.

 

[Reff]

Hi Dalespam and ghwellsjr
Thanks for your replys. Thanks for the challenge DaleSam but that formula is beyond me and probably not bring me to the point where I find where I am going wrong. My problem is that using the logic I have explained and a little geometry I get the correct SR time dilation figures for a moving frame. What I need to know is at what point does my logic break down.
This is why I am attempting to explain my logic perhaps with not altogether approved terminology.
ghwellsjr. Thanks for your question. Sentence 1) is closest but consider by reciprocal I mean it in a navigation sense in that for me it is obvious that a photon on a specific heading
must have a reciprocal return heading line going back to its origin this for me must be an imaginary line of logical thought. The reciprocal also happens to line up with the heading of the photon heading out in the opposite direction of the first photon. The reciprocals of the three photons I talk about all return to the same point that I have marked with a small object with a little mass. The three photons could also be reflected directly back and arrive back at the object. They left together and return together. I understand that propagation is independent of the frame they leave and photon speed of two photons heading in the same direction are identical irrespective of the speed and direction of the frames they leave.
I have no reference to the reciprocal heading of a photon, just my own logic. Making it up sounds a little harsh but if that is where I am going wrong I would appreciate knowing why.
I see two frame speeds which give rounded figures with geometry of the time dilation experienced by each. If you both use the correct formula and arrive at the same answers perhaps you would look at the geometrical logic for the same answers ie .6c frames clock dilates by .8 and a .8c frames clock dilates by .6.

 

[Dale]

by reciprocal I mean it in a navigation sense in that for me it is obvious that a photon on a specific heading
must have a reciprocal return heading line going back to its origin this for me must be an imaginary line of logical thought. The reciprocal also happens to line up with the heading of the photon heading out in the opposite direction of the first photon.

Even though it is not a standard term, this is how I understood it. The standard term for the path a particle takes through spacetime is "worldline", so you are just talking about looking back along the worldlines.

I reassert, in every frame the photons form a sphere centered on the origin expanding at c and in every frame any two non-colinear photon worldlines intersect at the origin. It is interesting that despite your objections on this point you still get the right answers fot time dilation. Unfortunately, I don't follow your description well, but the math is clear and unambiguous.

 

[Reff]

Hi DaleSpam
Let me give you a description of the geometry and perhaps you may see something in that.
Draw a 10cm radius circle which represents a propagating sphere of light.
Draw a line right through the centre of the sphere from the bottom to the top. This will represent the direction of a moving frame we wish to calculate time dilation for.
The frame is a table, the top which is flat face to the direction of travel. The left hand edge of the table is adjacent to and heading along the direction line.
We begin by creating an event next to the left hand edge of the tabletop and allow the sphere to expand to the 10cm radius.
In this example the frame is moving at .8c so we can place the left hand edge of the table 8cm up from the centre of the sphere, in the direction of travel.
Draw the table top from the 8cm point of the direction line at right angles to the direction line out to the right intersecting the sphere.
The table has intersected the sphere and I reason, there is a photon right there at the surface of the table and at the intersection. This photon like any other on the sphere has moved 10cm from the event at at the spheres marked centre at c. Its reciprocal is back to the marked event.
The intersection marks a particular photon which has moved during propagation at c, from the event to the intersection just like any other in the sphere. Now measure the distance the photon has traveled on the top of the table. It will be precisely 6cm. What is interesting is if you change the scale of the drawing, it is always the same photon that is crossing the tabletop so the tabletop observer will see a crossing photon moving 6cm and not its actual distance of 10cm so in measuring c he is using the frames time dilated clock and has no option but to mistakenly measure it moving at c. This crossing photon could never be overtaken by any other photon. If we use a clock made using a datum frequency from bouncing a photon from one side of the tabletop to the other it must move slower than c and be time dilated but with the bigger picture we see it moving at c just like the tablecrossing photon from the event to the tabletop intersection

 

[matphysik]

I assume in an SR example of a propagating sphere of light from a pinhead size event that its propagation symetry is independent of the frame which created it. The point on the frame which created the event must be somewhere within the sphere.

Use three photons on the sphere to create a target. 1. Any one photon 2. A photon heading out on the reciprocal of the first photon. 3. A photon heading out at 90 degrees to the line formed by the first two photons.

During propagation of the three photons take a small target of some mass and hold it on the reciprocal of the three photons for a short duration and release.
The target will now be inertial and remain on the reciprocal of the three photons.

Is it reasonable to suggest the photons move at c from the point marked by the target to the edge of the sphere but not at c from the point on the frame they left unless that point remains with the target.

Are these reasonable assumptions or at what specific point am I mistaken. I find this can be used as the basis for simple SR calculations.

Hello. In SR we must speak of light or a continuous beam of photons, not individual photons (whose behaviour is described by QM).

 

[Reff]

Hi Matphysik
Thanks for your correction. Yes I do have a problem with the correct terminology. You may see my intent by considering the rate of expansion of a propagating sphere of light being c on its radius. I must admit to considering a photon as a particle moving at the same rate of expansion as a spheres radius or a beam of light.
I submitted a new posting as yours arrived which probably includes more incorrect terminology but I trust you will have a read and check out how I reached my time dilation figure of .6t for a moving frame moving at .8c

 

[ghwellsjr]

I'm still having a lot of trouble trying to understand your scenario or what you are trying to do with it so let me ask you some questions to help me understand your points.

1. Are you aware that any inertial (non-accelerating) observer who sets off a flash of light will think, believe, and measure, that he is in the center of the expanding sphere of light?

2. Are you aware that a second observer inertially moving at a high speed relative to the first observer if he happens to be colocated with him at the moment the flash is set off, will also think, believe, and measure that he is in the center of the expanding sphere of light?

3. Is this the issue you are concerned about, how they can both think they are in the center of the expanding sphere of light even though they are no longer colocated?

 

[CJames]

Hi DaleSpam
Let me give you a description of the geometry and perhaps you may see something in that.
Draw a 10cm radius circle which represents a propagating sphere of light.
Draw a line right through the centre of the sphere from the bottom to the top. This will represent the direction of a moving frame we wish to calculate time dilation for.
The frame is a table, the top which is flat face to the direction of travel. The left hand edge of the table is adjacent to and heading along the direction line.
We begin by creating an event next to the left hand edge of the tabletop and allow the sphere to expand to the 10cm radius.
In this example the frame is moving at .8c so we can place the left hand edge of the table 8cm up from the centre of the sphere, in the direction of travel.
Draw the table top from the 8cm point of the direction line at right angles to the direction line out to the right intersecting the sphere.
The table has intersected the sphere and I reason, there is a photon right there at the surface of the table and at the intersection. This photon like any other on the sphere has moved 10cm from the event at at the spheres marked centre at c. Its reciprocal is back to the marked event.
The intersection marks a particular photon which has moved during propagation at c, from the event to the intersection just like any other in the sphere. Now measure the distance the photon has traveled on the top of the table. It will be precisely 6cm. What is interesting is if you change the scale of the drawing, it is always the same photon that is crossing the tabletop so the tabletop observer will see a crossing photon moving 6cm and not its actual distance of 10cm so in measuring c he is using the frames time dilated clock and has no option but to mistakenly measure it moving at c. This crossing photon could never be overtaken by any other photon. If we use a clock made using a datum frequency from bouncing a photon from one side of the tabletop to the other it must move slower than c and be time dilated but with the bigger picture we see it moving at c just like the tablecrossing photon from the event to the tabletop intersection

My brain hurt trying to visualize what you were describing, but I think I finally understand. What you have basically done is derived the origin of time dilation. You used a different setup, but you used the Pythagorean theorem to demonstrate how time dilation works, the same way Einstein did.

From the external reference frame, the table is moving at .8c. From this reference frame, the beam of light has traveled 10 cm from its origin, and it is now located 6 cm further along the table.

From the reference frame of the table, the light has only moved 6 cm.

What this means is that there has been enough time for light to move 10 cm in the external reference frame, but there has only been enough time for light to travel 6 cm in the table's reference frame.

This is why time dilation occurs in the first place.

Have I answered your question?

 

[Reff]

Hi ghwellsjr
Well thanks for your reply and thanks for the obvious effort you have gone to to try and understand my odd stuff.
Yes I am aware of the observers of both inertial frames believing each is the center of the spheres of light and that each believes they measure the speed of light at c.
If you have a go at my geometry answer to DaleSpam you will see how I derive the individual frames time dilation byusing geometry and making the observation that light crossing a moving frames table does not cross at c. This is my concern that I use my perhaps incorrect logic but get the correct maths.

 

[Dale]

Let me give you a description of the geometry ... a table, the top which is flat face to the direction of travel.

Why don't you draw it and post a drawing. For instance, "flat face to the direction of travel" I cannot tell if you mean that the direction of travel is along the face or normal to the face. We could clarify this line by line, but a drawing would get it all done at once.

the tabletop observer will see a crossing photon moving 6cm and not its actual distance of 10cm

There is no "actual" distance. There is only the distance in a given frame. The 6 cm in the tabletop frame is just as valid as the 10 cm in the other frame.

so in measuring c he is using the frames time dilated clock and has no option but to mistakenly measure it moving at c.

Similarly, the tabletop observer's measurement of c is not a mistake.

If we use a clock made using a datum frequency from bouncing a photon from one side of the tabletop to the other it must move slower than c

No, it is a postulate of relativity that it moves at c.

 

[ghwellsjr]

I'm still having a lot of trouble trying to understand your scenario or what you are trying to do with it so let me ask you some questions to help me understand your points.

1. Are you aware that any inertial (non-accelerating) observer who sets off a flash of light will think, believe, and measure, that he is in the center of the expanding sphere of light?

2. Are you aware that a second observer inertially moving at a high speed relative to the first observer if he happens to be colocated with him at the moment the flash is set off, will also think, believe, and measure that he is in the center of the expanding sphere of light?

3. Is this the issue you are concerned about, how they can both think they are in the center of the expanding sphere of light even though they are no longer colocated?

Hi ghwellsjr
Well thanks for your reply and thanks for the obvious effort you have gone to to try and understand my odd stuff.
Yes I am aware of the observers of both inertial frames believing each is the center of the spheres of light and that each believes they measure the speed of light at c.
If you have a go at my geometry answer to DaleSpam you will see how I derive the individual frames time dilation byusing geometry and making the observation that light crossing a moving frames table does not cross at c. This is my concern that I use my perhaps incorrect logic but get the correct maths.

I'm not sure if you answered my third question. I asked about a single sphere of light and you answered about spheres of light. Are you thinking that the light is behaving differently for each observer, that is, there is one expanding sphere of light for the stationary observer and a different expanding sphere of light for the moving observer?

 

[CJames]

Hi ghwellsjr
Well thanks for your reply and thanks for the obvious effort you have gone to to try and understand my odd stuff.
Yes I am aware of the observers of both inertial frames believing each is the center of the spheres of light and that each believes they measure the speed of light at c.
If you have a go at my geometry answer to DaleSpam you will see how I derive the individual frames time dilation byusing geometry and making the observation that light crossing a moving frames table does not cross at c. This is my concern that I use my perhaps incorrect logic but get the correct maths.

Reff,

I think you missed my post. I have already responded to your geometrical setup. What you have done is demonstrated why time dilation occurs in the first place, whether you realize it or not.

For everybody else, picture a circle with a radius of 10 cm. Draw a right triangle inside of this circle. One corner is at the center of the circle, one corner is 8 cm above the center of the circle, and one corner is on the circumference of the circle.

We know the left side is 8 cm up, and that the hypotenuse is 10 cm, so we can use the Pythagorean theorem to demonstrate that the top of the triangle is 6 cm across.

The top of this triangle represents the surface of a table after it has moved 8 cm from the origin. During this time, the light sphere has expanded by 10 cm.

From the external reference frame, the beam of light that has traveled along the hypotenuse of the triangle has moved 10 cm. But it has moved 6 cm across the surface of the table.

What Reff is apparently failing to realize is that he has demonstrated WHY time dilation occurs in the first place.

From the external reference frame, the beam of light has moved 10 cm. This means that 10cm/c time has passed.

From the reference frame of the table, the beam of light has only moved 6 cm. This means that 6cm/c time has passed.

Again, this is WHY time dilation occurs.

So hopefully Reff will read my response this time, and hopefully I have answered his question.

 

[CJames]

I'm not sure if you answered my third question. I asked about a single sphere of light and you answered about spheres of light. Are you thinking that the light is behaving differently for each observer, that is, there is one expanding sphere of light for the stationary observer and a different expanding sphere of light for the moving observer?

I think what he is saying is that from the external reference frame the light sphere has a 10 cm radius, and from the internal reference frame it is only 6 cm.

I think what's confusing him is that, from the external reference frame, it looks like the beam of light is traveling slower than c in comparison to the table. We avoid that language in SR to avoid that confusion.

 

[Reff]

Hi CJames and other posters.
I am unintentionaly a little behind with answering all the postings possibly because of my time zone and more answers than I bargained for. I have just overnighted and have one eye on a recording of le tour.
Sorry about your brain hurting CJames
Thanks for your input.
You have obviously followed my geometry. Consider the picture you have drawn and consider a question to the moving observer " where did the photon start from, and he and most people will say here--" and point to the left hand side of the table. I am saying That after frame movement and propagation of the sphere, the left hand side of the table is a point within the sphere where the crossing photon has never been. My consideration for the crossing photon is it is in a state of continuous intersection with the tabletop and it started from the same point as all the other photons in the sphere. It moves at c as with all the other particles. I am saying it therefore does not cross the tabletop at c but always at less than c in any moving frame.
I don't believe there is any geometry for SR because of the observers aberation of the point of origin of the crossing photon.
Using the described geometry try a variation. Reduce the frame speed relative to propagation and keep the frame table edge aligned with the point of origin. Now we have an interesting variation because a photon from the event can indeed cross the table. Its trajectory is indeed across the top of the table and only then is the crossing speed c and also its frame is not time dilated and is in fact reading absolute time and dare I suggest the frame is at, for the want of better words "absolute rest" .
Without googling perhaps I remember absolute rest being voted out by consensus some time in the past. I still believe the simple SR geometry still gives the correct answers and is still sound.
Consider this from ghwellsjr
1. Are you aware that any inertial (non-accelerating) observer who sets off a flash of light will think, believe, and measure, that he is in the center of the expanding sphere of light?

2. Are you aware that a second observer inertially moving at a high speed relative to the first observer if he happens to be colocated with him at the moment the flash is set off, will also think, believe, and measure that he is in the center of the expanding sphere of light.

If I use a little geometry on this, firstly yes I would agree with both observers believing they are both in the center of the expanding sphere of light and if they both had tables to measure the speed of light and the moment that light reaches the other side of the table they each create an event to mark the occasion then the geometry would show the markers to be separate to the center of the sphere.

 

[Reff]

H DaleSpam
Sorry I am slow and perhaps exasperating
I do have a little geometry but for the moment perhaps you could consider the table edge represented by the single straight line at 90 degrees to the direction of travel line. its left hand edge is on the direction of travel line. Take any table you see and imagine it taking off verticaly from the floor and that is my poor description of its direction of travel.
My other geometry gives two other slightly more detailed moving frame calculations I am not sure if it will copy and paste or what-- for this posting.

 

[Reff]

Hi ghwellsjr- Quoting
I'm not sure if you answered my third question. I asked about a single sphere of light and you answered about spheres of light. Are you thinking that the light is behaving differently for each observer, that is, there is one expanding sphere of light for the stationary observer and a different expanding sphere of light for the moving observer?

I believe both observers will see an expanding sphere of light centered on them but if we stay centered on all the the headings or reciprocals--world lines and observe the two frames, then the moving observer I believe is mistaken. A marking event from this observer will show up the aberation.
We should be able to center ourselves on any "event" and consider a photon on the edge of the sphere at each radius and a measurement of each radius should be identical. Whilst still observing the sphere, a frame which created it moving at say .9c would no longer be centered after any propagation and frame movement. The frame would need to reverse to be hit by all the photons if the photons reversed direction. I see light moving at c absolutely independently of the speed of the frame it leaves. That part of the frame it leaves must be within the sphere even if the frame is fast moving and close to the edge. The only no aberation frame to view this from is centered on the sphere and the center of it marks absolute rest-- yes this is purely my view for the present. My geometry works fine on with this

 

[Dale]

Sorry I am slow and perhaps exasperating

I wouldn't say either slow or exasperating, just confusing. You write many paragraphs, but I can't visualize the scenario you are trying to describe.

In the absence of a clear picture from you I will just have to point you back to the derivation I provided above. Whatever you may draw, in any frame the light always forms a sphere centered on the origin and the "reciprocals" always intersect at the origin. The math is firm on that point.

 

[CJames]

I wouldn't say either slow or exasperating, just confusing. You write many paragraphs, but I can't visualize the scenario you are trying to describe.

In the absence of a clear picture from you I will just have to point you back to the derivation I provided above. Whatever you may draw, in any frame the light always forms a sphere centered on the origin and the "reciprocals" always intersect at the origin. The math is firm on that point.

See my post, #19. Hopefully it is easier to visualize.

 

[CJames]

My consideration for the crossing photon is it is in a state of continuous intersection with the tabletop and it started from the same point as all the other photons in the sphere. It moves at c as with all the other particles. I am saying it therefore does not cross the tabletop at c but always at less than c in any moving frame.

Not in the reference frame of the table. From the point of view of the external frame, the beam of light is moving at c along the hypotenuse. At each point along this hypotenuse, you are correct in saying that the beam of light will be intersecting with the table. So from the external reference frame, it looks like the beam of light is traveling across the table at slower than c, but you will also observe that the beam of light's true speed is c.

When you start talking about the internal reference frame, however, things look different. Beams of light move at c in comparison with any observer. That is an axiom of special relativity, and one that has been experimentally verified in many circumstances. So from this perspective, the beam of light is not moving along the hypotenuse of a triangle. It is moving along a straight line across the surface of the table. It only travels 6 cm, instead of 10 cm. For this reason, less time has passed within this reference frame.

You could postulate that there is such a thing as "absolute rest," and claim that all the other reference frames are just getting the "wrong" answer because of the effects of time dilation and length contraction.

The problem with this is that there is no known way to decide which reference frame is "at rest." The laws of physics are the same in all reference frames. There is no known experiment that you could perform which would tell you that you are either "moving" or "at rest."

 

[Reff]

Hi CJames and DaleSpam
I am slightly overpowered with facts. but still hanging on.
I hope you have the geometry I have described sorted DaleSpam.
Just a basic thought I believe my geometry addresses a moving frames clock which is time dilated and that frames observer will always measure his light speed at c. I am fine with that but in any at rest frame observing that frames calculation, we find he is measuring a slow photon. Tha fact that we have trouble finding the at rest frame does not mean it is not so. History would show many thought experiments later verified by fact.
Quote CJames
So from this perspective, the beam of light is not moving along the hypotenuse of a triangle. It is moving along a straight line across the surface of the table. It only travels 6 cm, instead of 10 cm. For this reason, less time has passed within this reference frame.
Just consider this
Look closer at this beam of light and look at the photon make up. I believe we need to consider the release of each photon particle being radialy emitted from the at rest target and as the moving frame progresses another particle is emitted from a new at rest target point. This is a continuous process and the statement I have in mind is yes you can have a lateral beam of light which is not at rest orientated but every single photon will be emitted from its own at rest frame. So I believe all light waves from a moving frame have a lateral wave but all particles within it are -radialy- emmitted from their own event observed by an absolute rest observer. I presently happy with that I believe, so is my geometry.

Surely a photon particle is absolutely independent of any moving frame. It is not given extra speed or slowed down. The frame obviously cannot overtake it but for the observer to say " this is where the photon left from and point to the left hand side of his table-- no I still cannot see that, because the one photon we are talking about left from it own at rest frame which has been left behind by the moving frames speed.

 

[Drakkith]

Reff, your example works whether the observer emitting the light is moving or whether it is the other observer that is moving. That is the whole point. There is no absolute frame that we can say is at rest. To keep things simple we usually refer to the frame of the Earth as being at rest, but it is not.

Just a basic thought I believe my geometry addresses a moving frames clock which is time dilated and that frames observer will always measure his light speed at c. I am fine with that but in any at rest frame observing that frames calculation, we find he is measuring a slow photon.

Slow photon? What is that? To the observer moving it travels at c, and to the observer stationary it travels at c as well.

So from this perspective, the beam of light is not moving along the hypotenuse of a triangle. It is moving along a straight line across the surface of the table.

You don't look at it from the perspective of the moving observer, but of the stationary one. That is why it looks like it is traveling along the hypotenuse. The effect is much better demonstrated in the light clock example. When you have a moving observer with light bouncing being emitted, bouncing off a mirror, and being detected at the bottom, to that observer the light is moving only up and down. (Or whatever direction it is oriented.) When we observe that clock from a frame that isn't moving, the light is no longer going straight up and down, but is traveling in a diagonal, aka the hypotenuse of a triangle. The faster that clock travels in relation to the other frame, the longer the hypotenuse is. See below.

Observer in the rest frame of the clock.
http://en.wikipedia.org/wiki/File:Time-dilation-001.svg
Observer in a rest frame where the clock is moving.
http://en.wikipedia.org/wiki/File:Time-dilation-002.svg

I believe we need to consider the release of each photon particle being radialy emitted from the at rest target and as the moving frame progresses another particle is emitted from a new at rest target point. This is a continuous process and the statement I have in mind is yes you can have a lateral beam of light which is not at rest orientated but every single photon will be emitted from its own at rest frame. So I believe all light waves from a moving frame have a lateral wave but all particles within it are -radialy- emmitted from their own event observed by an absolute rest observer. I presently happy with that I believe, so is my geometry.

I'm not sure what you mean by this. Each photon travels in a straight line in both the stationary and the moving observers frame. However in the stationary observers frame the photons emitted in the direction of travel are Blue shifted and have more energy, while the ones away from the direction of travel are Red shifted and have less energy.

Surely a photon particle is absolutely independent of any moving frame. It is not given extra speed or slowed down. The frame obviously cannot overtake it but for the observer to say " this is where the photon left from and point to the left hand side of his table-- no I still cannot see that, because the one photon we are talking about left from it own at rest frame which has been left behind by the moving frames speed.

Lets not confuse frames here. The frames only matter for the observers, the photons have no frame themselves. You can easily say that X photon was emitted from Y position on an object. Otherwise how would we see anything moving at all?! The non moving observer seeing an image of, (what was it again, a table?), would see the table as warped and length contracted depending on the direction of travel in relation to the observer, but you could still tell it was a table at lower % speeds of c.

 

[Reff]

Thanks for your reply Drakkith
Interesting
Re absolute rest. I must subscribe to the belief that absolute rest exhists although there is no experiment to suggest it does exhist. I believe as I have said before that it was voted out by consensus by a group of scientists.
Why I say this is simply because two photons heading in opposite directions have a start point and if they return from their heading after each has completed a specific distance they
will arrive at the same time. This I believe is the frame of the photon and that point would have to be absolute rest. As you say, for now, the Earth will do. When I talk of photons---particles I am not referring to a beam of light but the leading particle. They are instantly up to c whatever the frame speed they leave.
Think about the fastest time clock in the universe and it must be co located at an absolute rest frame. There could be an infinate number of these clocks but none could ever meet because they would have to time dilate to meet another.
I will have a go at the rest of your post after I have sorted a little mowing and bits..

 

[Drakkith]

Thanks for your reply Drakkith
Interesting
Re absolute rest. I must subscribe to the belief that absolute rest exhists although there is no experiment to suggest it does exhist. I believe as I have said before that it was voted out by consensus by a group of scientists.

That is what happens when things are shown to be incorrect. They become obsolete and aren't used anymore. This has happened countless times in the past, why is absolute rest any different than anything else before it?

Why I say this is simply because two photons heading in opposite directions have a start point and if they return from their heading after each has completed a specific distance they
will arrive at the same time. This I believe is the frame of the photon and that point would have to be absolute rest. As you say, for now, the Earth will do. When I talk of photons---particles I am not referring to a beam of light but the leading particle. They are instantly up to c whatever the frame speed they leave.

Again, photons do not have a frame. When we talk about a frame, we refer to an Inertial Frame. One where the object is at rest. A photon CANNOT have a rest frame, it always travels at c. For example, our two observers are passing by each other at 50% c. Either one can say that the other is at 50%c, as both views are correct. But what if one of these is light? An object with mass cannot travel at c, so for a photon it CANNOT say that it is stationary and you or I are traveling at c. Get what I'm saying?

Think about the fastest time clock in the universe and it must be co located at an absolute rest frame. There could be an infinate number of these clocks but none could ever meet because they would have to time dilate to meet another.
I will have a go at the rest of your post after I have sorted a little mowing and bits..

This doesn't even make sense.

 

[Dale]

we find he is measuring a slow photon.

No such thing.

every single photon will be emitted from its own at rest frame.

Also, no such thing.

observed by an absolute rest observer

If there is such a thing nobody knows, including the lucky absolute rest observer. It is a physically useless concept, but feel free to use it if it makes you feel better. Lorentz aether theory is experimentally indistinguishable from SR, and basically consists of doing normal SR and occasionally throwing in a transformation to a frame labeled "aether" which is moving with an unknown velocity.

Why I say this is simply because two photons heading in opposite directions have a start point and if they return from their heading after each has completed a specific distance they
will arrive at the same time. This I believe is the frame of the photon and that point would have to be absolute rest.

Then according to this all frames are absolute rest.

 

[ghwellsjr]

Hi ghwellsjr- Quoting
I'm not sure if you answered my third question. I asked about a single sphere of light and you answered about spheres of light. Are you thinking that the light is behaving differently for each observer, that is, there is one expanding sphere of light for the stationary observer and a different expanding sphere of light for the moving observer?

I believe both observers will see an expanding sphere of light centered on them but if we stay centered on all the the headings or reciprocals--world lines and observe the two frames, then the moving observer I believe is mistaken. A marking event from this observer will show up the aberation.
We should be able to center ourselves on any "event" and consider a photon on the edge of the sphere at each radius and a measurement of each radius should be identical. Whilst still observing the sphere, a frame which created it moving at say .9c would no longer be centered after any propagation and frame movement. The frame would need to reverse to be hit by all the photons if the photons reversed direction. I see light moving at c absolutely independently of the speed of the frame it leaves. That part of the frame it leaves must be within the sphere even if the frame is fast moving and close to the edge. The only no aberation frame to view this from is centered on the sphere and the center of it marks absolute rest-- yes this is purely my view for the present. My geometry works fine on with this

OK, in this and other posts, you have indicated that you don't have a correct understanding of what's going on with an expanding sphere of light. I have created a series of animations to illustrate and hopefully help you understand what's happening.

First of all, you have indicated that you think that an absolute rest state, one that has been described as an ether, is the only one in which light travels at c. That's fine, we'll start with that idea and later I will show that it is identical to selecting any arbitrary inertial frame of reference as defined in Einstein's Theory of Special Relativity. So let's pretend that there exists an absolute ether rest state in which light travels at c in all directions and only in that state does light travel at c in any direction.

Now imagine an observer at rest in this ether and he sets off a flash of light. Wouldn't everyone agree that this will form an expanding sphere of light, enlarging at the speed of light, with the observer in the center? (We won't concern ourselves with the individual photons, we'll assume that the light is so bright that it creates a wavefront of light traveling away from our observer.) Now we have to ask ourselves the question: how does the observer decide if he is in the center of this expanding sphere of light? He cannot see the light that is traveling away from him, can he? It's gone away from him.

Well, the easiest way for him to figure out if he is in the center is to place a bunch of mirrors equally spaced from himself and wait for the sphere of light to hit the mirrors and reflect back to him. If he sees the reflections from all the mirrors arrive back to him at the same time, he can validly deduce that he was in the expanding sphere of light. In our diagrams and animations, we get to watch the wavefront of the light both while it is expanding and after the reflection, while it is collapsing, but we have to understand that the observer in the center of all this activity only knows of the initial emission of the light and its final reception, both of which happen at his location.

Now an animation to illustrate an expanding sphere of light in three dimensions would be very difficult to implement, but we can easily demonstrate the salient points by limiting it to a plane with an expanding circle of light, a circle of mirrors, and a contracting circle of reflected light. In the video that follows, the observer is shown as a green stick man with the light emitted as a blue circle from the center of his round head. He has placed a full circle of yellow mirrors all around him at an equal distance from his round head. The expanding circle of blue light reflects off the circular mirror as a collapsing circle of green light. Watch this:



Pretty simple, isn't it?

Next we want to consider another observer who is traveling at one-half the speed of light and see what happens to him. He also will place what he thinks is a circle of mirrors some equal distance around him. But it turns out that due to length contraction along the direction of his motion, his mirrors actually form an ellipse as shown in this animation:



Here, the moving observer is shown as a contracted red stick man and the reflection of the circle of light is also in red. Do you see how the light always travels at c no matter whether it is traveling away from him or reflected back towards him? Do you also see how the elliptical shape of the mirrors actually transplants the center of the expanding circle of light to a new location where the collapsing circle of light ends up?

The thin black dashed line shows the locus of points where the two partial circles of light contact the mirror and can be used to illustrate the reflections that occur in a light clock, although, usually a light clock is shown with just the horizontal sections of mirror and a light flash traveling up and down between them.

Finally, in order to show both stick men at the same time, they need to make their mirrors be partial, otherwise one of them will collect all of the reflected light leaving none for the other one:



Notice how the reflected light first collapses on the green stationary man and then later on the red moving man. This illustrates the time dilation experienced by the moving man.

Now it turns out that as far as the men can determine, they each are having exactly the same experience. The moving man not only concludes that he is in the center of the expanding circle of light, he also believes, and has every reason to believe, that the light struck every part of his mirrors at the same time and that his mirrors formed a perfect circle. Furthermore, he believes that the speed of light is a constant c in all directions relative to him because that is what he measures and even though he cannot see the progress of the light, it behaves exactly as if he were stationary in an absolute ether rest state.

It also turns out that any Frame of Reference that is defined according to the requirements of the Theory of Special Relativity will look exactly like these same animations--you cannot tell any difference between them and the absolute ether rest frame of the Lorentz Ether Theory.

 

[CJames]

Just consider this
Look closer at this beam of light and look at the photon make up. I believe we need to consider the release of each photon particle being radialy emitted from the at rest target and as the moving frame progresses another particle is emitted from a new at rest target point. This is a continuous process and the statement I have in mind is yes you can have a lateral beam of light which is not at rest orientated but every single photon will be emitted from its own at rest frame. So I believe all light waves from a moving frame have a lateral wave but all particles within it are -radialy- emmitted from their own event observed by an absolute rest observer. I presently happy with that I believe, so is my geometry.

As I have already said, you can arbitrarily choose the external reference frame to be at "absolute rest." It works. It just doesn't have any physical meaning in any known way.

In both reference frames, the beam of light started at the left side of the table. In both reference frames, the beam of light is now located 6cm across the surface of the table. In one reference frame, the table was moving, and the light has traveled 10 cm. In another reference frame, the table was at rest, and the light has traveled 6 cm.

There is no known scientific way to choose which reference frame was at rest. It is completely arbitrary.

If you insist on saying that one reference frame is at rest, you can still get the right answers. It looks like this is what you have done. Maybe it is simply easier for you to visualize it that way. I just hope you aren't insisting that there has to be a reference frame at absolute rest, because nothing you've portrayed here indicates that there must be.

 

[Reff]

Hi ghwellsjr and other posters
Well thanks for your last post. The videos are brilliant. I need a little time here to digest them, as you may have noted I am more of a procrastinator. I had a realisation overnight re my geometry of the .8c frame dilation which was explained in clearer terms by CJames, that readers are believing that I am on about a beam of light and not a particles, I will explain. If we go back to the .8c geometry and I am not discounting yours ghwellsjr. If we use a beam of light and allow it to form a sphere of light 200mm dia then it is obvious to most that the beam is still being generated at the scource in any frame.The trajectory is straight across the table which would seem to make me incorrect. If we take frames at any speed we have the same effect. All frames experience the same as you have impressively demonstrated.
With my geometry, it is a single particle on a constant state of intersecting the tabletop and not a beam of light. If that photon returns straight back on its heading, then the frame must reverse to meet it. If the frame was at absolute rest then the photon will return to that frame and that frame remains inertial. The beam of light is a different scenario. If the beam has moved 100mm, it is still being generated at the left hand side of the tabletop which logicaly one would take as a point of origin but, consider a constant stream of photons within the beam as being generated from a constant new point of absolute rest.
There is no length contraction for my geometry as the tabletop lies in the lateral plane and the maths work fine. There is no pythagarus required as the 200mm dia circle and the associated relativity are precisely to scale.
Quoting Drakkith re absolute rest.
That is what happens when things are shown to be incorrect. They become obsolete and aren't used anymore. This has happened countless times in the past, why is absolute rest any different than anything else before it?

If Absolute rest has been shown to be incorrect then I would be incorrect, but I don't believe that to be so because was it conveniently voted out not by observation but by consensus.

CJames had a good understanding of the geometry although he had problems with it.
I would ask that consideration be given to a single particle crossing a tabletop and its point of origin, because it is not a beam of light which is still being transmitted during propagation and which would return to the light generator, now the particle has an origin at the confluence of all the particle headings now allow the frame to progress a further cm and create the next event, which now becomes another photon in a state of constant intersection with the tabletop. Now constanly release particles while the moving frame continues on and the constructed beam would appear to be crossing smoothly across the table with the beam starting at the event generator, but every single particles heading will go back to absolute rest.
Was it Einstein who said the " Aether must be the nature of a solid body, because transverse waves are not possible in a fluid, but only in a solid"
What I believe is All photons are radialy emitted from absolute rest and form transverse waves appearing to come from any moving frame.
Does QM have some incompatability issues with SR and perhaps GR.

 

[CJames]

If that photon returns straight back on its heading, then the frame must reverse to meet it. If the frame was at absolute rest then the photon will return to that frame and that frame remains inertial.

How do you determine whether or not a photon is returning straight back on its heading? There is no known experiment you can perform to prove that this is what is happening.

If Absolute rest has been shown to be incorrect then I would be incorrect, but I don't believe that to be so because was it conveniently voted out not by observation but by consensus.

The point is, the idea of absolute rest ceased to have any observable predictions. What scientific use is a theory that doesn't make predictions?

Does QM have some incompatability issues with SR and perhaps GR.

Quantum field theory reconciles quantum mechanics and SR. Quantum mechanics is currently incompatible with GR. It is every theoretical physicist's dream to figure out how to reconcile the two.

 

[Reff]

Hi CJames
I do feel I am fighting a losing battle but still see the clarity in my observation and geometry. I thought about the "slow photon" several years ago and eventualy decided to put a little maths to it and crudely reached the required figures eventualy using c as 1 and transposing a 3 4 5 triangle to reach the rounded figures. It works with any speed. I started with GR initialy and still reached the "slow photon" geometry but the maths for that is way above me and the geometry still tricky. I would find it hard to concede having worked out SR calculations for myself.

Sure I feel a photon has a heading-- not found yet but the right experiment perhaps. If a photon can hit an electron over a distance-- sure it has a heading.

Absolute rest observable predictons not found yet but yes some smarty will devise prediction experiments and if found it gives a new meaning to relativity in general for instance as I have just said, radial construction of tranverse waves and the acceptance of an aether.


Quantum mechanics being incompatable with GR is interesting, With incompatability comes the chance of reconciling the two. So who is wrong. I would put my money on GR.
If space was classed as a liquid with transverse waves being formed by continuous intersection of the lateral frame and not a curved solid perhaps then perhaps we can class gravity as a pure flow of space and again the transverse photon moves slower than c.
Perhaps that reconciliation is closer than we think.

 

[Reff]

Hi Bessie 11
I am not a research person. I have an interest in anything geometrical. For many years I have puzzled over GR and SR and now feel I have a reasonable view of the phenominum.
I am sort of retired and on occasion make weird things! sort of inventions. If you wish to see a couple try utube search under "VAWT with rudder". There are a couple of bits there and perhaps more eventualy.
As for SR calculations they are easy to work out geometricaly but not easy to convince others how I arrived at the answers. For the moment the geometry of relativity is not going too far. GR geometry was a little more accepted.

 

[Dale]

If Absolute rest has been shown to be incorrect then I would be incorrect, but I don't believe that to be so because was it conveniently voted out not by observation but by consensus.

This is incorrect. It is more than a century of the most sophisticated observation possible: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

At this point, a belief in an absolute rest frame is on par with a belief in the Tooth Fairy. It cannot be ruled out in principle, but it doesn't appear to actually do anything since all available data can (more) easily be explained without it.

 

[Dale]

For the moment the geometry of relativity is not going too far.

Complete BS. Modern relativity is all about geometry. You should learn about the following:
Spacetime diagram
Minkowski norm
Spacetime interval
Four-vectors
Tensors
Riemannian geometry

 

[Reff]

Hi DaleSpam
You could be right.

 

[CJames]

Hi CJames
Sure I feel a photon has a heading-- not found yet but the right experiment perhaps. If a photon can hit an electron over a distance-- sure it has a heading.

But not an absolute heading. In one reference frame that heading is across the surface of the table. In another it is from a point where the table used to be. (I also feel the need to say that when you're talking about a photon, instead of "a pulse of light," you run into quantum mechanical issues, where the photon doesn't have a precise location or momentum.)

Absolute rest observable predictons not found yet but yes some smarty will devise prediction experiments and if found it gives a new meaning to relativity in general for instance as I have just said, radial construction of tranverse waves and the acceptance of an aether.

Lorentz ether theory makes the same predictions as SR, except that it arbitrarily chooses one reference frame as the reference frame of the ether, which only serves to complicate matters without adding any additional observable predictions.

Quantum mechanics being incompatable with GR is interesting, With incompatability comes the chance of reconciling the two. So who is wrong. I would put my money on GR.
If space was classed as a liquid with transverse waves being formed by continuous intersection of the lateral frame and not a curved solid perhaps then perhaps we can class gravity as a pure flow of space and again the transverse photon moves slower than c.
Perhaps that reconciliation is closer than we think.

GR doesn't postulate that space is either a liquid or a solid. Not sure what you're getting at here?

 

[Reff]

Hi CJames.
I feel shortly I will be consigned to the Crackpot drawer but for now
Quoting you
But not an absolute heading. In one reference frame that heading is across the surface of the table. In another it is from a point where the table used to be. (I also feel the need to say that when you're talking about a photon, instead of "a pulse of light," you run into quantum mechanical issues, where the photon doesn't have a precise location or momentum.)
This is interesting and in fact as I was about to have a browse into QM, your post came up.
Your geometry would seem to be strong and you have understood most of my intent even if you disagree re the reference frame of the crossing photon. I believe that it is --for the want of better words-- the frame of the photon which regulates everything. It is the geometry of that to me which is the king frame. A photon particle in free flight has its own datum speed. what effect has a frame on c after the photon has left. For sure it has an effect on a beam of light which must eminate from the frame but not the particle.
From an event just look at two photons leaving in opposite directions -- the moving frame has gone-- they are synchronised together and move at 2c relative to each other. They each can impart energy to a distant electron and so have a heading and a path reversal of that is back to a meeting point which has remained inertial throughout. Any frame which has to accellerate back in the opposite direction to meet that point is moving in my books. You may not agree but consider the path of the photon as being laser straight until it encounters gravity so in open gravity free space a single photon can hit a distant electron. It is also tidy to make that distant electron part of an absolute rest frame and all is precise. The heading of a photon towards a distant absolute frame, will hit that frame.
I once read a guys post re the impossibility of measuring the speed of light. He could be right but a reconciliation with QM would be interesting in that regard.
For sure I am frustrating but my geometry bas based on absolute rest for years and the maths agreeing sort of makes it hard to let go.
Quote
GR doesn't postulate that space is either a liquid or a solid. Not sure what you're getting at here?
At the risk of changing direction. I believe SR is an induced flow by frame speed and the lateral crossing photon is slow. With GR it is the flow of space where absolute rest is being taken into gravity so a frame on Earth has similar geometry to SR and the crossing photon is still slow due to its absolute rest frame having been taken in by the flow. My analogy is a launch in a river directly crossing the flow so its crossing speed is less than its hull speed.
I am sure you will be able to draw the geometry for that. I convinced a respected poster of that years ago who now uses a canoe as his analogy.
Quote
GR doesn't postulate that space is either a liquid or a solid. Not sure what you're getting at here?
The man himself said If space is a liquid then transverse waves are not allowed. If we have absolute rest and transverse waves are formed from the continuous emission of photons from a continuous movement of the photon at rest frame, then transverse waves can be formed from the process.
Thats geometry for a headache. Hard to follow if I have no credibility.
Lorentz has some interesting stuff I have not touched on for a while

 

[Dale]

I believe that it is --for the want of better words-- the frame of the photon which regulates everything.

Again, there is no such thing. https://www.physicsforums.com/showthread.php?t=511170 This is not even something that is open to the "tooth fairy" wiggle room of the idea of an absolute frame. Even if there is an absolute rest frame it is not the frame of a photon because such a concept is not even logically self consistent.

 

[CJames]

I believe that it is --for the want of better words-- the frame of the photon which regulates everything. It is the geometry of that to me which is the king frame. A photon particle in free flight has its own datum speed. what effect has a frame on c after the photon has left. For sure it has an effect on a beam of light which must eminate from the frame but not the particle.

As DaleSpam has said, and you seem to partially acknowledge, there is no "frame of the photon." I'm sure you're referring to a hypothetical electromagnetic field that is "stationary."

From an event just look at two photons leaving in opposite directions -- the moving frame has gone-- they are synchronised together and move at 2c relative to each other. They each can impart energy to a distant electron and so have a heading and a path reversal of that is back to a meeting point which has remained inertial throughout. Any frame which has to accellerate back in the opposite direction to meet that point is moving in my books.

I'm going to insist on talking about pulses of light and mirrors instead of photons and electrons, because of the quantum mechanical effects that would severely mangle this thought experiment.

So imagine two pulses of light travel outward from an origin point, hit two mirrors, and return to the origin point.

Now suppose that this setup was on the table in your previous thought experiment. The experiment would play out exactly the same way, and it would be the external reference frame that would need to "accelerate back in the opposite direction to meet that point."

For sure I am frustrating but my geometry bas based on absolute rest for years and the maths agreeing sort of makes it hard to let go.

I can understand that, but insisting on an absolute reference that has no observable predictions only leads to confusion in my opinion.

As for the rest of your post, it appears more philosophical than scientific. Whether you imagine space as a liquid or a solid, all we can really say is that it is something we can measure using speed and time.

 

[Reff]

Hi Dalespam and CJames
Thanks for your replys.
Could I just go back to what CJames calls a pulse of light. I presume it is reasonable to suggest a pulse of a very short duration say its width 1cm and its length 1cm and is also a parrallel- non convergant or divergant light. Now from the event which creates the pulse we watch two pulses of heading in opposite directions. Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again
If the pulse was from a .8c frame are you saying the reflected pulse will go back to the pulse generator on that frame or somewhere else because that frame has moved on.
That point between two pulses of light can be used to create more pulses in sequence which will remain centered on all the others irrespective of the speed and directions of the frames they leave. What possible effect can frame speed and direction have on a light pulse which has left it. It is free to move by its own frames laws the moment it leaves
Two pulses-- and I prefer particles close on the same trajectory from different frames, What does it matter where their frames have moved to. One photon cannot overtake another.
I have suggested before, that I have geometry to my formula whilst your formula does not re DaleSpam. How about showing me how you would do your SR calculations using geometry including vectors.
How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

 

[ghwellsjr]

Hi Dalespam and CJames
Thanks for your replys.
Could I just go back to what CJames calls a pulse of light. I presume it is reasonable to suggest a pulse of a very short duration say its width 1cm and its length 1cm and is also a parrallel- non convergant or divergant light. Now from the event which creates the pulse we watch two pulses of heading in opposite directions. Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again
If the pulse was from a .8c frame are you saying the reflected pulse will go back to the pulse generator on that frame or somewhere else because that frame has moved on.
That point between two pulses of light can be used to create more pulses in sequence which will remain centered on all the others irrespective of the speed and directions of the frames they leave. What possible effect can frame speed and direction have on a light pulse which has left it. It is free to move by its own frames laws the moment it leaves
Two pulses-- and I prefer particles close on the same trajectory from different frames, What does it matter where their frames have moved to. One photon cannot overtake another.
I have suggested before, that I have geometry to my formula whilst your formula does not re DaleSpam. How about showing me how you would do your SR calculations using geometry including vectors.
How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

Reff, if you would relate what you are saying to my animations, I think it could help us understand what you are trying to say, since you have not produced your own diagrams.

My first animation seems to correspond to your scenario concerning two pulses of light that head in opposite directions and are reflected back to their point of origin, except that I have a whole bunch of pulses of light going in every possible direction, but you could draw a line through the head of the green stick man and that would correspond to the headings of your two pulses. They go out and hit a portion of the mirrror on opposite sides, simultaneously and reflect back and collapse on the man at the same time. Doesn't that fit with what you are saying for the first part?

Then for the next part, you talk about a .8c frame but I used a .5c "frame" except I hate to call it a frame because we're still looking at it from the same frame as the first situation. What we have now is another observer who is traveling at .5c and when he gets to the point where the first man was, the same flash of light is emitted in all directions. It's the same flash as in the first case but this time the moving man carries along with him his own set of mirrors and when the light hits his mirror, it reflects back differently than it did for the first man who was stationary in the frame. And when it reflects back it hits the moving man at a new location.

You can do the same thing I suggested for the first part, draw a line through the moving man's head at the point where the flash is emitted and pretend like there are just two pulses of light going in opposite directions, but they travel relative to the stationary frame, not relative to the moving man. They eventually hit the moving man's mirror and you have to draw a new line from the point of contact to the point where the man will be when he finally gets to the point where all the pulses of light collapse simultaneously on his head.

I drew the black dashed line so that you would recognize it as an ellipse with the two foci corresponding to the starting point of the flash and the collapsing point of the flash and I'm sure you're aware that every line going from the first focus point to a point on the ellipse and back to the second focus point has the same total distance as any other line. This is a requirement of the light traveling at a constant speed in the stationary frame, not in the moving frame.

I cannot understand what you mean here:

That point between two pulses of light can be used to create more pulses in sequence which will remain centered on all the others irrespective of the speed and directions of the frames they leave. What possible effect can frame speed and direction have on a light pulse which has left it. It is free to move by its own frames laws the moment it leaves​

And the reason this doesn't make any sense is because you keep talking about light pulses leaving frames. What does that mean? Whenever you're dealing with a scenario like this, you must use just one frame for everything. All the light pulses travel at c in that one frame. It doesn't matter how they got emitted, by a source stationary in the frame or a source moving at a high speed in the frame. Light pulses don't leave frames.

And I can't make any sense out of the rest of your post.

So could you please relate the two situations of an observer at rest in a frame to my first animation and an observer moving with respect to a frame in my second animation and see if you agree with what they are depicting. If not, what do you find objectionable with them?

 

[Dale]

Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again

They have a heading in any reference frame. The heading is frame-dependent. When a pulse of light hits a reflective surface it obeys the laws of reflection in all frames.

I have suggested before, that I have geometry to my formula whilst your formula does not re DaleSpam. How about showing me how you would do your SR calculations using geometry including vectors.

Certainly. For reference see the following links which describe the geometric elements used here:
http://en.wikipedia.org/wiki/Four-vector
http://en.wikipedia.org/wiki/Minkowski_space

I will use a convention where capital letters are for four-vectors, and X \cdot Y denotes the Minkowski inner product of the four-vectors X and Y.

A sphere of light emitted at t=0 from the origin is given by all events X such that:
X \cdot X = 0

If \eta is a Lorentz transform then it preserves the Minkowski inner product such that in another reference frame we have
X&#039; = \eta X
X&#039; \cdot X&#039; = X \cdot X = 0

So in any other frame the events also form a sphere of light emitted at t=0 from the origin.

How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

Reff, there is no physical significance to the concept of absolute rest to the best experimental accuracy possible to date. This has been demonstrated over and over and over again during the course of the last 100+ years of increasingly sophisticated and accurate experimental tests. See: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html . The fact that you are unaware of the experimental evidence against your position does not invalidate that evidence.

Your geometrical reasoning is also incorrect. This was demonstrated by me in post 7. The fact that you are unable to follow the algebra does not invalidate the math.

You need to spend less effort trying to justify your mistake and more effort learning the relevant scientific and mathematical concepts that you are missing. Once you have done so then you will either understand why you are wrong or you will have the tools to defend your position rationally.

 

[CJames]

Now from the event which creates the pulse we watch two pulses of heading in opposite directions. Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again

As Dalespam said, they have a heading but that heading is different in each reference frame. Yes, they can be reflected directly back to meet again. All I'm saying is that you could place the two mirrors on a table that is "moving" at .8c, or you could place the two mirrors on the floor which is "at rest," and in both scenarios the light would be reflected back to the origin.

What looks like two pulses of light moving away from the origin are returning to it in one reference frame, looks like two pulses of light moving away from one another, and meeting again at some other point in space.

If the pulse was from a .8c frame are you saying the reflected pulse will go back to the pulse generator on that frame or somewhere else because that frame has moved on.

Within any reference frame, if there are two mirrors and a source of light bisecting them, two pulses of light will reflect off of the mirrors and return to the source of light simultaneously. It doesn't matter if that frame is "at rest," because all frames are at rest wrt themselves.

What possible effect can frame speed and direction have on a light pulse which has left it.

Redshift/blueshift.

It is free to move by its own frames laws the moment it leaves
Two pulses-- and I prefer particles close on the same trajectory from different frames, What does it matter where their frames have moved to. One photon cannot overtake another.

Nobody is saying one photon can overtake another. Light pulses don't have their own frame.

How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

It is not traveling at c wrt itself, it's traveling at c wrt everything else. The fastest clock is in any inertial reference frame, measured from within that reference frame.

 

[Reff]

Hi ghwellsjr
thanks for replying again.
Yes your animations are really very good although I don't necesserily agree. Now Re the first video yes I totaly agree with this one provided that it represents an inertial frame centered on the propagation. Now I did explain my geometry of SR earlier on for a frame moving at .8c and this was further clarified by CJames I believe. Now if you draw my geometry and I presume you agree it gives the correct answer for .8c frame and any other frame speed you care to name. Draw a 200mmdia circle and draw a vertical line right through the middle of it now this is your first animation. The circle is as far as propagation has reached. Your inertial frame is at the center. Photons are heading in all directions but we are only interested in one photon or a small pulse of light which is heading along the surface of the tabletop. When it hits a reflector at the end of the tabletop, it returns straight back on the surface of the table to its point of origin. that pulse of light or particle or photon is moving at c and back from the source.
Now do my .8c diagram and note the frame speed compered with the vector of the photon the edge of the table is following-- it is marked off simply at .8c. That is where the left hand edge of the tabletop is .8c up from the event. Now draw a photon vector from the event to where the tabletop intersects the sphere. There is a photon on the edge of the sphere and it is on the surface of the tabletop. That very same photon has been on the surface of the tabletop right from the event up to where it is now. It has moved .6 of the propagation we are looking down on so without any maths at all we know the moving frame is .6time dilated because it is simply a transposed 345 triangle. I can do this for all frame speeds and it works fine. The .6 time dilation is simply happening because it is actualy moving at c like absolutely any other photon in the radius off the sphere, they are all there to see. BUT what is the table crossing cspeed .6 it is a little tricky to understand because it is not crossing the table top it is constantly intersecting the tabletop. Big difference.
Let's go back to your first diagram. If we put a tabletop in your diagram the left hand side of this is where your little animated man is- as I have said-- its light progresses at c and returns at c Is there a vector to the crossing photon---no-- is there any time dilation--- no--because it crosses at c. has the left hand side of the tabletop moved up the vertical line No. his is a clock that in my books reads absolute time. There is no faster clock than this.
If you do my geometry as described and overlay yours as I have described then you can perhaps understand how I derive my figures purely geometricaly.
Your animation would be interesting to show how a moving frames tabletop crossing photon crosses in a state of constant intersection and the intersection speed of an inertial moving frame is always less than c. The point of origin of the photon and the constant intersection could be hilighted.
Its tricky geometry. Just ask yourself, if a clock stops at c, there must be a clock that runs faster and faster with a decreases in frame speed. The fastest is centered on your first diagram.

 

[Reff]

Hi CJames and DaleSpam
You have both given me plenty of information to get on with. Thanks for the effort. I would have to be more than a little obstinate to try and convince you guys otherwise. Blind and cannot see. perhaps I will have a look at QM,there would appear to be conflict of theory there. For sure I believe I am right but I understand to go on will mean contravening the agreement to post plus I now fancy a beer. If this remains open I may just answer your posts on a nicer day.