1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

SR Question

  1. Apr 7, 2004 #1
    An atom is at rest in an inertial reference frame. Suddenly, two photons are emitted from it, at right angles to one another.

    What is the relative speed of the photons?

    P.S. Assume they depart from the atom at 299792458 meters per second in the atomic frame, as has been found true by experiment.
    Last edited: Apr 7, 2004
  2. jcsd
  3. Apr 7, 2004 #2
    SR Answer: c
  4. Apr 7, 2004 #3
    Prove that the relative speed of the two photons is c if you can.




    By Euclidean geometry, if the relative speed of the photons is c, then the two photons form an equilateral triangle with the atom, whose area is increasing in time. The interior angles of an equilateral triangle are 60 degrees. However, the emission angle was stipulated to be 90 degrees. This is a contradiction.
    Last edited: Apr 7, 2004
  5. Apr 7, 2004 #4
    What exactly allows you to apply Euclidean geometry to spacetime in this case? (I am sincerely asking this because I don't know the answer.)
  6. Apr 7, 2004 #5
    Answer: The Pythagorean Theorem

    Keep in mind that the time dilation formula of the special theory of relativity is derived by assuming that the pythagorean theorem is true in the kind of reference frame we have in my question here.

    By stipulation, the photons are departing from the atom at right angles to one another in the rest frame of the atom, which is an inertial reference frame by stipulation. The distance between them at any moment in time is found using the pythagorean theorem. We actually have an isosceles triangle here. The hypotenuse of which is greater than either leg.

    However, if the relative speed of the photons is c, as was suggested by Jdavel, then you can prove that the triangle is equilateral. Thence, the emission angle of the photons is 60 degrees.

    But it was stipulated that the emission angle is 90 degrees.

    Kind regards,


    P.S. The very first theorem of Euclid, was to prove that an equilateral triangle could be constructed. Here is a link to Euclid's first theorem:

    Euclid 's First Theorem

    In Euclid's eleventh proposition, he proves that a right angle can be constructed:

    Euclid Book One Theorem Eleven
    Last edited: Apr 7, 2004
  7. Apr 7, 2004 #6
    First Issue:
    I posed this question to StarThrower in post #92 on this page of the following thread:

    I proposed that the photons would never observe each other and therefore could never determine how fast they were traveling relative to each other. An observer (in this case the atom) at point C could certainly propose how fast they were traveling with respect to himself, or how fast they were traveling with respect to each other with respect to himself. I'm intrigued about the legitimate answer.

    Second issue:
    Flat space is only one kind of space. As you say, "By stipulation, the photons are departing from the atom at right angles to one another in the rest frame of the atom." But you cannot necessarily assume that the photon is restricted to Euclidian 3-space. The Pythagorean theorem applies to flat space. For curved spaces you must use a modified version of the theorem that takes into account the metric of the curvature of space.


    The metric is a (sometimes complex) matrix that tells you HOW to measure the distance between any two points in a specific locale of the space.
  8. Apr 7, 2004 #7
    Let me reiterate:

    All kinds of mathematical nonsense is resting on the incorrect assumption that speed of light is c in any inertial reference frame. Space is three dimensional Euclidean.

    From the site quoted we have:

    In Euclidean three-space, the distance between points A=(x1,y1,z1) and B=(x2,y2,z2) is:

    [tex] D(A,B) = \sqrt{ (x2-x1)^2 + (y2-y1)^2+(z2-z1)^2 } [/tex]

    You cannot bend the vacuum.

    Kind regards,

    Last edited: Apr 7, 2004
  9. Apr 7, 2004 #8
    You're in the wrong profession, mate. You should have become a comedian.

    How come light, which travels at straight lines, can and does bend around the sun, allowing us to see objects which are actually behind it? Or are you going to suggest that light does not travel at a straight line now?
  10. Apr 7, 2004 #9
    How do you define gravity in 3-space?
  11. Apr 7, 2004 #10
    For light to travel in a curved path, it must experience a force. Otherwise, by Newton's First Law of Motion (Which is roughly Aristotle's law of Inertia), the photon would continue to move in a straight line at a constant speed.

    BUT YOU ARE TELLING ME THE PHOTON IS IN A GRAVITATIONAL FIELD. Therefore, something is interacting with the photon to exert a force on the photon. Hence the photon isn't in an inertial reference frame anymore.

    Kind regards,

    Last edited: Apr 7, 2004
  12. Apr 7, 2004 #11
    Like I said define the force of gravity in FLAT 3-space.
  13. Apr 7, 2004 #12

    [tex] \vec{F} =- GM_1 M_2 \frac{\vec{R}}{|\vec{R}|^3} [/tex]
  14. Apr 7, 2004 #13
    Indeed it was. According to this equation when you wave your hand from left to right (specifically when you modify R in the equation above), you instantaneously influence atoms on jupiter, the moon...etc, because your hand exerts a gravitational force on them. Are you comfortable with this?
  15. Apr 7, 2004 #14


    User Avatar
    Homework Helper

    Space-time does NOT have a Euclidean geometry; it has a Euclidean topology. Since the pathagorean theorem is a geometrical theorem for Euclidean geometry, it does not apply to space-time.
  16. Apr 7, 2004 #15


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Undefined, since, as we've pointed out to you before, photons don't have rest frames.

    Allow me to remind you that the difference of two velocity vectors in a given reference frame is not, in general, give you the relative velocity of the said objects in SR.

    According to GR, the photon is travelling in a straight line, period. And, of course, photons never have inertial rest frames.
  17. Apr 7, 2004 #16

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    After all the time people have spent showing you that a reference frame moving at c is ill-defined in SR, I can't believe we're seeing another "what does motion look like from the photon's point of view?" thread.

    So what??

    The angle is stipulated to be 90 degrees in the lab frame. Even in Galilean relativity, the angle would not be 90 degrees in any other frame. In fact, it is always possible to transform to a frame in which both photons are moving along the same line! It's called the "center of momentum" frame.
  18. Apr 7, 2004 #17
    Hurkyl, is this because in order to determine the relative velocity of two objects they must be in inertial rest frame(s) (IRF)?

    Let me pose a sub-light question (god forbit)...two objects A and B speed away from point C. At time [tex]t_{0}= 0[/tex] the distance between any two of these three points (A, B, and C) can be considered zero. A and B are traveling at 0.7c wrt point C.


    How do we determine how fast A and B are travelling with respect to each other? Usually in SR this kind of thing is stated up front (S' is traveling at speed v with respect to S).

    StarThrower asserts (as he usually does) that we can state how fast A and B are traveling with respect to each other if we just use our observations from our own frame of reference and apply [you know who's] theorem. I don't believe this is correct. If we concentrate on sub-light speeds for this problem perhaps we can get "real" SR answers instead of having to listen to StarThrower assert something else about absolute time and space...
    Last edited: Apr 7, 2004
  19. Apr 7, 2004 #18

    I can prove it. But I can't prove it to you, because your mind is closed to the possibility that I can. So I'm not going to waste my time.

    A few days ago I showed you the source of all your confusion here: you don't understand what the variables used in SR mean. You think you do. But you don't.
    Last edited: Apr 7, 2004
  20. Apr 7, 2004 #19


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, I do interpret "relative velocity" to mean the velocity of one object in the rest frame of the other object.
  21. Apr 7, 2004 #20


    User Avatar

    Staff: Mentor

    On what do you base this assertion? And don't say the Pythagorean theorem - I want emperical evidence, for example, give me just one (confirmed) test of the speed of light that shows a variation from C.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: SR Question
  1. Sr & Gr (Replies: 3)

  2. Question about SR (Replies: 30)

  3. Quaternions and SR (Replies: 20)