"Light is frame-dependent in PF, but constant in SR"

Discussion in 'Physics & Math' started by Maxila, Jan 22, 2013.

  1. Maxila Registered Senior Member

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    Below is a quote in the OP from the thread “Comparison of Special Relativity with a Galilean "preferred frame" theory” by poster James R, which brought up an issue I wanted to discuss. I thought this was sufficiently off that topic to warrant a new thread:
    As I’ve looked at SR, the constant speed of light postulate has become more intriguing because it appears implicit in SR that c, in a way, can be considered a constant, yet a frame dependent constant? I know there are posters that are substantially more advanced in physics than I am, so perhaps if I outline how I interpret it they can help me understand where I may have erred?

    As far I know the constant for c is the ratio of distance per time or x/t (i.e. 299,793,458 per second or 299,793,458/1). Using this, and information on the Muon Experiment from http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html I will explain why it appears implicit to me that x/t of c is “observed” constant (or invariant) from every frame, yet it is also relative constant value to a frame?

    Note, as per the site I referenced, t is stated in microseconds, muon speed observed from the ground is .98c, gamma is 5, and I will round c to .3 kilometers/1µs (microsecond), unless otherwise noted.

    Essentially I can convert the ground frame’s x and t to the muon frame simply by multiplying them by the inverse of the gamma factor 5 (1/5). Where the ground clock measures the muon traveling for 34 microseconds the muon clock records 34*1/5 or 6.8, the ground observes a distance of 10km the muon observes 10*1/5 or 2km.

    Where I see this invariant, yet frame relative c, is that for each frame they must see c as .3km/1 µs (x/t), yet we also know that x/t as seen from the muon frame is (x/t)/5 as compared to the ground frame, implying a ground value of c = .3km/1 µs, when the muon value is c = (.3km/1 µs)/5 for the muon?

    This does makes sense how c is invariant as seen from a frame, yet also a frame specific value, because where the ground will see light travel 1okm from the muon in 33.333us, the muon will see that light travel 2km in 6.666 µs. In other words the muon observes 1/5, the inverse of the gamma factor, for the light distance and time (x/t); implying that for the muon in order for that frame to observe c as .3km/1us relative to its x/t, that value must also be 1/5 of the grounds c?

    My use of (?) is to indicate I know my reasoning may be flawed, yet the math is simple, I am certain it is correct, and the deductions seem sound and logical. There are a few other things that also seem implicit in the math also.

    I. That the muon must see x in any direction (x. y, z) as x/5 relative to the ground frame
    II. Space (x, y, z,) in the muon frame must be contracted by x/5 relative to the space of the ground in order for both of them to observe a constant and invariant value of c relative to each frames value for x and t to c = x/t.

    It would be most helpful if any responders referred directly to this example when discussing any errors or flaws I may have made in my reasoning.

    Maxila
     
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  3. Syne Sine qua non Valued Senior Member

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    Frame-dependence means that the value varies according to the velocity of that frame, so as the velocity changes so does the measured speed of light. Since the value c is invariant in all frames, regardless of velocity, the standards of time and space, as observed of other frames, must differ.
     
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  5. brucep Valued Senior Member

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    It's not a constant in EVERY frame. The local coordinate speed of light is invariant. The remote coordinate speed of light

    dr/dt = 1-2M/r

    Can be any speed between 0 and c. It doesn't have anything to do with 'seeing'. It's about measurements. Gamma doesn't figure into any measurements of the speed of light. "1/5 of the grounds c" is nonsense. You're trying to say the local coordinate speed of light is relative. It's not.
     
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  7. eram Sciengineer Valued Senior Member

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    Sorry, what is this remote coordinate speed of light?
     
  8. Robittybob1 Banned Banned

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    Could someone please explain in simple terms what "the remote coordinate speed of light" means.
    An example will help so I get the picture of what is being measured. Thanks
     
  9. brucep Valued Senior Member

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    4,098
    Th local coordinate speed of light is measured in a frame where the effects of gravity can be ignored. A segment of lights path where the geometry of spacetime is essentially flat. This local segment, where the effects of gravity can be ignored, exists for a segment of the path for all objects following the natural path through the gravitational field.

    The remote measurement has to account for the effects of gravity over the entire path. For example the remote coordinate speed of light at r=2M, the event horizon, is

    dr/dt = 1-2M/r=1-2M/2M=1-1=0c

    or at r=4M

    dr/dt = 1-2M/4M=1-.5=.5c

    When r=very large the remote coordinate speed of light approximates the local measurement but it's not invariant.
     
  10. Robittybob1 Banned Banned

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    If R = 4M does that mean M is a length? What sort of length are we talking about?
     
  11. brucep Valued Senior Member

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    4,098
    It's in geometric units like I explained to you 3 or 4 times at the other site. 1 solar mass equals 1477 meters or 1 solar mass or any length or any amount of solar mass. What ever you choose since it doesn't make any difference to the solution. This isn't my thread so don't troll me with a thousand questions.
     
  12. Maxila Registered Senior Member

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    156
    That is what I thought I demonstrated; only I added (what appeared as a logical interpretation of that math), that since time and space x/t is the inherent the value of c, than in order for c to remain a constant in every frame it has to change with x/t? Clearly above the muon sees a 1/5 value relative to the ground for x/t so the muons c must also be 1/5 the grounds c, otherwise it would disagree with the values of x/t the muon observes (it wound not be seen as .3km per microsecond it were not also 1/5 the grounds, relative to the 1/5 value of x/t the muon observes)?

    It does make sense of how everything is observed as frame dependent, also of how c is still a frame constant (frame invariant).

    Maxila
     
  13. Syne Sine qua non Valued Senior Member

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    No, you are conflating the frames of reference involved. All observers have a local, invariant standard of time and length. They only witness non-standard times and lengths in remote frames, i.e. in motion relative to the observer.

    Overall, you are not making much sense.
     
  14. RJBeery Natural Philosopher Valued Senior Member

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    Yes, Maxila, we agree on this. The phenomenon you're describing would explain Lorentz transforms and local c invariance. I would challenge the others to come up with a logically consistent alternative explanation for how c could remain constant for all local frames.
     
  15. brucep Valued Senior Member

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    Like Syne said you're conflating frames. I wonder how many times I need to say this? The frames where the speed of light is invariant is local coordinate and local proper frames. The speed of light is frame dependent when measured from remote coordinate frames. The reason your logic is irrelevant is because you don't understand this basic part of the theory. Why the measurements in local coordinate and local proper frames are invariant. This science is useless to anybody who can't understand that. Events in spacetime are invariant because the measurements describing spacetime events are made where the event occurs. Not from remote coordinates. Measurements made from remote coordinates are frame dependent. I suggest that you use a text to learn the theory. This internet attempt hasn't worked to well for you.
     
  16. RJBeery Natural Philosopher Valued Senior Member

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    You aren't providing a reason why they are locally invariant, you're just claiming they are by fiat and then saying that we don't understand physics (which is quite an odd comment); Maxila and I are proposing a reason why they are locally invariant.
     
  17. Maxila Registered Senior Member

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    That's excatly what I demonstrated with the example and math. It's straightforward and amazing if you can visualize the math in an empirical setting.

    Maxila
     
  18. brucep Valued Senior Member

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    No you didn't.
     
  19. Syne Sine qua non Valued Senior Member

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    Just another hack seeking refuge in an argument from consensus by commiserating with other hacks. This really should not be allowed in the Physics & Math forum.
     
  20. RJBeery Natural Philosopher Valued Senior Member

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    OK you need to knock this crap off. Keep it civil and if you don't appreciate the logic then just move on. The ad hominems don't make you guys looks any smarter.
     
  21. Maxila Registered Senior Member

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    156
    Bruce and Syne:

    I gave a specific example with calculations for a reason. So a responder could point exactly to what is incorrect and what they disagree with. State specifically what are the problems you see within the OP; otherwise you are spewing words and demonstrating you don't understanding anything.

    *Science requires (*note this is a science forum) such a response for your reply to have any merit. The data in the example came from the Georgia State University physics department and the link was provided in the OP. Your reply is not credible without stating facts and supporting evidence directly addressing the OP, until such is provided your opinion remains scientifically pointless and unsubstantiated.

    Maxila

    P.S. I'm more than willing to constructively discuss any specific errors you point too...
     
  22. brucep Valued Senior Member

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    You don't understand anything I say. You just brush it aside to continue to believe what you wrote is correct. One more try. Let's say the 'muon' can measure the local coordinate speed of light over a segment of it's path. In the muon local coordinate frame, defined by the segment of the path where the effects of gravity can be ignored and where the measurement is made [flat spacetime], the measurement IS INVARIANT. No matter where this path is in our universe the measurement IS INVARIANT. Even if the muon has fallen into the black hole and it's proper velocity is > c. The same: IE a constant. It's exactly the same for all local coordinate and local proper frame measurements of the speed of light. IE a constant. The same. Remote coordinate speed of light measurements are frame dependent. They're frame dependent because the measurement takes into account the effects of gravity, spacetime curvature, over the entire path. If you have a calculation that concludes the muon measures the local coordinate speed of light at 1/5 c. It's wrong. It's you playing with mathematics that describe natural phenomena that you don't understand.

    I'll just add this about the Shapiro Delay. If we measure the local coordinate speed of light at any point over the path the measurement is invariant. Measured over the entire path from remote coordinates there's a delay.

    The light leaves our sun and is received on our planet.

    The remote radial coordinate speed of light

    dr/dt = (1-2M/r)

    You can integrate this over the path and build a formula for predicting the Shapiro delay. For our case

    dt = [r_earth orbit - r_sun] + 2M_sun ln[r_earth orbit/r_sun]

    The 1st component is the distance between where the light is emitted and received and the 2nd component is the predicted Shapiro delay due to a path through curved spacetime.

    I'll solve it for our case

    r_earth orbit = 1 AU = 1.495978E11 m

    r_sun = 6.9598E8 m

    M_sun = 1477 m [using geometric units]

    dt_bkkp = [ 1.495978E11m - 6.9598E8m] + 2954m ln[1.495978E11m/6.9598E8m]

    = 1.4890182E11m + 15854.11645m = 1.489018359E11m

    15854.11645m is the delay time in geometric units. dt_meter.

    To convert this to seconds divide by c.

    dt_second = 15854.11645m/2.99792458E8m/s

    = .00005288367752 second

    ~ 53 microsecond

    That is the extra time due to the path through curved spacetime.
     
  23. OnlyMe Valued Senior Member

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    Brucep, you seem to do this quite often, confuse the theory with proof. By the emphasis you place, with bold type, you move the argument from one of argueing the theory, to one of fact. The fact is your statements are consistent with theory and have not been proven to be true... At least if they have been you have offered no reference to the experiments, measuring the velocity of light.., say sufficiently distant from "our" locally flat inertial frame, to support your insistence that the velocity of light is invariant in all imertial frames EVERYWHERE in the universe.

    Don't get me wrong I am not challenging the theoretical basis, just what seems to be an insistence that what is theory, be accepted as having been proven. Were that really the case there would be no reason to conduct any number of experiments. With cesium clocks and satellites, demonstrating relativistic time dilations (GPS satellites in orbits in the neighborhood of 26,000 Km), has anyone actually even attempted to measure the velocity of light with a clock known to be time dilated? And yes if we can assume a lab here on earth represents a flat spacetime for the purpose of the measurement of the speed of light, then the same could be assumed for any inertial frame.., but that remains theoretical, until proven.

    So, since you are so insistent on the UNIVERSALLY INVARIANT stance, I am sure you can provide a credible reference to the experimental measurement of the velocity of light in an inertial frame sufficiently distant from the surface of the earth to acts even a limited proof.
     

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