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Two beams emitted from single source

  1. Jan 4, 2016 #1
    Hello,

    This is my first post and I have one question.
    My major is not physics. For now, I am struggling to understand general relativity for own reason because the math involved is troublesome to me but facing one question I need to figure out.

    It is related with the gravitational lensing effect. I made a picture to help you understand my question better.

    2312F440568A60242CFE1A.png

    I heard that Einstein's general relativity provide the most correct picture about the geometrical curvature of space-time and mass is the source to warp fabric space-time. So light emitted from distant galaxy always moves straight line on the flat surface of space-time in vacuum but straight line on the warped surface of space-time in gravitational field near sun.

    My question is that is there any change in the frequency or wavelength of beam bent within gravitational field near sun?

    Let me describe morel from the picture above. Lets assume that a ship with double gun laser far away from sun and single laser source with frequency f1 at position p1 shoot laser beam through two gun at the same time to the direction of sun such that two beams bend slightly differently and passing through p2 for upper gun and p3 for lower gun. The path from p1 to p2 is equal length with the path from p1 to p3. The energy of two beams emitted from the ship is ##E = hf_1 ## .

    With such preparations, let suppose two observers are positioning at p2 and p3 and their watches are synchronized already such that they can start and complete measurement at the same time. Time synchronization is required due to the gravitational time dilation. Their task is to measure the frequency of the light beam from the ship.

    Now here is my question. What is the relation between f2 and f3? Are they same or different? If different, which one is higher? What general relativity predict? Is there any experimental results? My best guess is what general relativity predicts would be f2=f3 but not sure.


    I would appreciate for any help or guide.
     
  2. jcsd
  3. Jan 4, 2016 #2

    PeterDonis

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    No; GR predicts f2 < f3. But the reason has nothing really to do with gravitational lensing; it simply has to do with gravitational blueshift--light that is "falling" into a gravitational potential well blueshifts as it falls, and the deeper it goes, the more it blueshifts. So since p2 is not as deep in the Sun's gravity well as p3, the laser light that "falls" to p2 will blueshift less than the light that "falls" to p3. Blueshifting increases the frequency, so less blueshift means lower frequency, i.e., f2 < f3. (But both f2 and f3 will be larger than f1.)

    I suspect the question you really meant to ask is whether the lensing process itself, i.e., the different bending of the light beams passing through p2 and p3, produces any frequency change. AFAIK the answer to that is no.
     
  4. Jan 4, 2016 #3
    There's one common clock the observers use, and that causes them to measure same frequencies for the two beams.

    If two clocks are forced to be in sync all the time, there's really just one clock.

    Let's say the upper clock is the master clock. In this case the lower observer might say: "Light blueshifts by 10% when traveling through potential difference between upper observer and me, and I'm forced to use the upper observers clock which is 10% too fast."
     
    Last edited: Jan 4, 2016
  5. Jan 4, 2016 #4

    pervect

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    How do you synchronize the clocks at p2 and p3 if they are at different gravitational potentials?

    It appears to me that p2 and p3 are at different distances from the sun from the diagram, if they are they have different gravitational potentials. This implies that you cannot actually synchronize the clocks if the clocks keep proper time.

    I'm not quite sure of the intent of your question - are you thinking that p2 and p3 both use a coordinate time? Or are you thinking that p2 and p3 use proper time, i.e. a non-adjusted atomic time per the SI standard? I'd tend to assume the later, but then there is no need to even talk about synchronizing the clocks, as clock synchronization is not needed to measure proper time, so I'm not sure if that's what you had in mind.

    If we assume the clocks are measuring proper time according to the SI standard (in which case there is no need to synchronize them), f2 and f3 are different if and only if the gravitational potentials at f2 and f3 are different. They appear different on the diagram, I'm not sure if we have sufficient information to rigorously calculate the gravitational potential.

    If we assume the clocks are not measuring proper time, but that the raw measurements of proper time are being converted into coordinate time, we need to specify the details of the coordinate system. Most likely the coordinate system will be specified such that f2 and f3 have the same frequency, but it really depends on the details of the specification of coordinate time, and your written description is a bit sketchy so we may be imagining differing scenarios.
     
  6. Jan 4, 2016 #5

    PeterDonis

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    I was taking "synchronize" to mean only that the definition of simultaneity was the same for both p2 and p3, not that the elapsed time between corresponding simultaneous events was the same for both p2 and p3. The latter is, as you note, impossible if both p2 and p3 keep proper time. But the former is all that is necessary to formulate the question, because all we really want to enforce is that the two laser beams arrive at p2 and p3 at simultaneous events according to the common simultaneity convention of the two clocks (or more precisely, that successive wave crests of the two laser beams arrive at p2 and p3 at simultaneous events--but the elapsed times between the two crests, according to the clocks at p2 and p3, will be different, since those are the inverses of the measured frequencies f2 and f3).
     
  7. Jan 4, 2016 #6
    Ok, I missed that. But in this case, what I am still wondering is that the blue shifted beams of f2, f3 higher than f1 means that the beam energy are increased by the amount of h(f2-f1) and h(f3-f1) and the additional energy come from the gravitational field of sun. My basic understanding is that the energy of light is proportional with the measurable frequency.


    Yes, that was exactly what I meant to ask. Is the answer verified by experimental result or just prediction only? I am searching for both the prediction of GR and the verified result. Any similar verification would be good.


    I don't know exactly how to do it. My intention was thinking just about the definition of simultaneity. Maybe the simultaneity at p2 and p3 could be achieved by signaling from the control center. If someone press a button signaling the start of measurement, then that signal may arrive at p2 and p3 simultaneously and they start measuring. I hope this answer you.
     
  8. Jan 4, 2016 #7

    PeterDonis

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    That's true, but it's also important to understand that energy is frame-dependent; it's not the same for all observers. (Frequency is not either; an observer in orbit around the Sun, or free-falling towards the Sun, and passing by points p2 or p3 would measure a different frequency for the light from the laser cannon than observers at rest at p2 or p3.) When people say that light "gains energy" as it falls into a gravity well, what they really mean is "the energy of the light as measured by observers at rest is larger at lower altitudes".
     
  9. Jan 4, 2016 #8
    It makes me to think further about the photoelectric effect. Lets replace two observers with two same metals and one more metal near the ship to absorb the beam from additional light tube and all three metal plate's angle are adjusted to be equal to the incoming beam direction. If the beam is violet light, all metal plates absorbing beam emit electrons. What about the relation of the emitted electron's kinematic energy? If I read you correctly, the kinematic energy of electron at p3 in higher gravitational potential will be greater than the electrons emitted from the metal plate at p2 in lower gravitational potential.

    It also lead to the case of black hole singularity. Black hole can absorb light. As an photon is getting closer to the center of black hole, its position will be located in higher gravitational potential. Your reply forces me to think that a photon could produce humongous collision impact at the center of blackhole. Am I right or wrong?
     
  10. Jan 4, 2016 #9
    We have no idea of what is represented physically by the mathematical singularity showing up for idealised black holes.
    What is an infalling photon (or anything else) going to collide with?
    ... and whatever it is, will the classical physics of collisions even be applicable?
     
    Last edited: Jan 4, 2016
  11. Jan 4, 2016 #10

    PeterDonis

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    p3 is at a lower gravitational potential, not higher. (Think of gravitational potential as altitude--p3 is at a lower altitude than p2, hence a lower potential.) But yes, the UV photons received at p3 will have higher frequency, hence more energy, hence will cause the emission of photoelectrons with more energy (as measured by an observer at rest at p3), than the UV photons received at p2 (as measured by an observer at rest at p2).

    No. The concept of "gravitational potential" is only well-defined outside the hole's horizon. It does not make sense at or inside the horizon.

    Also, for an object that free-falls into a black hole, the energy that gets added to the hole (i.e., the increase in the hole's mass) is the energy the object has at infinity. The fact that it gains energy as it falls, according to observers "hovering" at lower and lower altitudes above the horizon, does not mean it gains energy as it falls from the standpoint of what energy gets added to the hole's mass.

    In short, black holes are counterintuitive in a number of ways, and you can't reason about them using the simple implicit model you are trying to use.
     
  12. Jan 4, 2016 #11
    It is my confusion. The fact that the light lose its velocity and staying in unknown form inside blackhole makes me to think the concept of colliding with something hard.


    Oops. I used the concept of gravitational potential differently. I understood that it was proportional with the slope of curvature of spacetime defined in general relativity or gravitational force( proportional inverse square of distance).
     
  13. Jan 4, 2016 #12

    PeterDonis

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    Light does not "lose its velocity" inside a black hole. Observers measuring the speed of light locally inside a black hole will measure it to be ##c## (just like anywhere else).

    No. Gravitational potential, in regions of spacetime where it is well-defined, is not the "slope" of anything. In the Newtonian approximation, where the concept of "gravitational force" is well-defined, the inverse square component of the force is proportional to the slope of the potential. The potential itself can be thought of, in this approximation, as just the "depth" of the gravity well.

    Also, spacetime curvature is not the same as "gravity" without qualification; that term can have a number of meanings. Spacetime curvature is the same as tidal gravity. In the Newtonian approximation, tidal gravity can be thought of as the difference in the "gravitational force" between two nearby points; but that is only an approximation and it breaks down in regimes like the region close to a black hole's horizon (or inside it).
     
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