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SR Photon target

  1. Jul 15, 2011 #1
    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.
  2. jcsd
  3. Jul 15, 2011 #2


    Staff: Mentor

    Hi Reff, welcome to PF!

    You mean place the target at the origin?

    I don't know what you are saying here.
  4. Jul 15, 2011 #3
    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.
  5. Jul 15, 2011 #4


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    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.
    Last edited: Jul 15, 2011
  6. Jul 15, 2011 #5


    Staff: Mentor

    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.
  7. Jul 16, 2011 #6
    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. Lets 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.
  8. Jul 16, 2011 #7


    Staff: Mentor

    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:
    [tex]c^2 t^2 = x^2 + y^2 + z^2 [/tex]

    Applying the Lorentz transform ( http://en.wikipedia.org/wiki/Lorentz_transformation ) in the standard configuration we get:
    [tex]c^2 \gamma ^2 \left(t'-\frac{v x'}{c^2}\right)^2=\gamma ^2 \left(x'-v

    Which simplifies to:
    [tex]c^2 \left(t'\right)^2=\left(x'\right)^2+\left(y'\right)^2+\left(z'\right)^2[/tex]

    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.
  9. Jul 16, 2011 #8


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    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.
  10. Jul 16, 2011 #9
    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.
  11. Jul 16, 2011 #10


    Staff: Mentor

    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.
    Last edited: Jul 16, 2011
  12. Jul 16, 2011 #11
    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 travelled 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
  13. Jul 16, 2011 #12
    Hello. In SR we must speak of light or a continuous beam of photons, not individual photons (whose behaviour is described by QM).
  14. Jul 16, 2011 #13
    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
  15. Jul 17, 2011 #14


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    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?
  16. Jul 17, 2011 #15
    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?
  17. Jul 17, 2011 #16
    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.
  18. Jul 17, 2011 #17


    Staff: Mentor

    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.

    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.

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

    No, it is a postulate of relativity that it moves at c.
  19. Jul 17, 2011 #18


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    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?
  20. Jul 17, 2011 #19

    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.
  21. Jul 17, 2011 #20
    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.
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