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B Do gravity and acceleration alter the speed of light?

  1. May 21, 2017 #1
    I am aware that an observer in free fall in a gravitational well will measure the local speed of light as c, but what about someone outside the well.

    Imagine this,

    Me and my friend are near Earth, and he starts freefalling to the surface, while i stay in space. If he then fires a laser beam at me, will I measure the laser travelling at c? And if I fire a laser at him, will he measure the laser travelling at c as well?

    Now imagine we are in space (no gravity) and we are moving at a constant velocity. Now imagine that I turn on a rocket engine and accelerate away from him, when he fires a laser at me, do I measure it as c, and if I fire at him, does he?

    Thank you for your help, Ivan.
     
  2. jcsd
  3. May 21, 2017 #2

    Orodruin

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    Unless you want to start introducing simultaneity conventions, all measurements of the speed of light are local and will give the same result. If you start introducing simultaneity conventions, your quantity will be a coordinate speed and not an actual speed.
     
  4. May 22, 2017 #3
    What do you mean by coordinate speed and actual speed? I thought speed were a coordinate dependent quantity i.e. that there were no means of talk about a speed without introducing a coordinate system.
    What do you mean by local? The two observers using the same coordinate system?
     
  5. May 22, 2017 #4

    Nugatory

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    The answer here comes from the answer to the next question.
    "Local" means within a region of spacetime small enough that we are unable to detect the curvature effects so can treat the region as flat. Within such a region the speed of light will be ##c## in all inertial coordinate systems, and because that works for all inertial coordinate systems we can call it an "actual" velocity, as opposed to a mere coordinate velocity which will change according to our choice of coordinate systems.
     
  6. May 23, 2017 #5
    Can you give a explicit example of an expression for a speed that changes with the coordinate system?
     
  7. May 23, 2017 #6

    Nugatory

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    Speed, defined by ##v=\frac{dx}{dt}## clearly is going to change when different coordinate systems assign different ##x## and ##t## values.

    Classical examples (so I don't have to mess with the relativistic velocity addition):
    My speed using coordinates in which the chair I'm sitting in is at rest at the origin is zero. Using the (non-inertial) coordinates in which the center of the earth is at rest at the origin my speed is about 400 meters per second. Using coordinates in which the jet plane flying overhead is at rest, my speed is something else again. And so on......

    This is all fairly natural in classical physics and special relativity. The transforms (Galilean or Lorentz) allow us to convert these speeds from one coordinate system to another in an unambiguous way, so there's always an answer to the only question that ever really matters, namely "How fast is Nugatory moving relative to <whatever>?" However, this cannot be done in the curved spacetime of general relativity, because answering that question requires parallel transporting my velocity vector to the origin of the coordinate system in which that <whatever> is at rest.... And in curved spacetime parallel transport is path dependent. Thus, if the curvature effects cannot be ignored there is no way of attaching any physical meaning to the coordinate velocity of a distant object, or a flash of light.

    And that is why we only talk about the local speed of light and why we define "local" as "near enough that curvature doesn't matter". Special relativity is actually the special case in which spacetime is flat, so the locale can encompass the entire universe.
     
  8. May 23, 2017 #7
    Thanks. Why can't we use the information we have about the path to parallel transport the vector to the origin in which the "whatever" is at rest?
     
  9. May 23, 2017 #8

    Orodruin

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    Which path to the origin?
     
  10. May 23, 2017 #9
    I don't know which one. It's the one that @Nugatory mentioned in his post.
     
  11. May 23, 2017 #10

    Nugatory

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    We do use that path information. It's part of the parallel transport calculation (Schild's Ladder is one example of how this might be done).

    The problem is that we get different results when we do the transport over different paths in curved spacetime; that's what "and in curved spacetime parallel transport is path dependent" means. There is no reason to choose one of the many possible paths as more correct than the others, so no way to declare any one of the different results to be the right one.
     
  12. May 23, 2017 #11

    Orodruin

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    His point was that the result would be path dependent.
     
  13. May 24, 2017 #12
    Ah, ok. Can you say a bit more on why measuring velocity in curved spacetime is not valid? I see your point that a vector in curvilinear coordinates is path dependent, but why do we need to transport it to the origin of our coordinate system to get a result?

    Suppose I am in a curved space (time) and I'm interested in measuring the spatial velocity of a particle. So what I need is just to find out the spatial position of that particle and to take the derivative of that position with respect to time. Also, as I'm in my own reference frame, the time appears to me just as a parameter, i.e. the time in the derivative is the one measured by my clock. So apparently in this process there are no need to transport the vector to the origin of my coordinate system.

    Is my current understanding wrong? Why?
     
  14. May 24, 2017 #13
    Or was you meaning that in curved spacetime both ##e_\mu (x)## and ##V^\mu (x)##, where the former are the basis vectors and the latter the components of a tangent vector ##V##, will depend on the point ##x##? If so, why can't we just use covariant derivative and other tools we have to evaluate that vector ##V(x) = V^\mu (x) e_\mu (x)## in another point ##x' \neq x##?
     
  15. May 24, 2017 #14

    PeterDonis

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    That's not what Orodruin and Nugatory are saying. They are saying that, if you have two events, A and B, in a curved manifold, and you have a vector at A, which vector at B you end up with when you transport the vector at A to B depends on the path between A and B that you transport it along.

    For a simple example of this, consider the following example on a 2-sphere, the surface of the Earth. Suppose I have a vector at the North Pole (point A), and it is pointing along the Greenwich meridian. I parallel transport it down to the equator along the Greenwich meridian; the intersection of the Greenwich meridian and the equator is point B. I end up with a vector at point B that points due south.

    But suppose instead that I parallel transport my vector at A down the 90 degree west meridian to the equator, and then along the equator to point B. Now I end up with a vector at point B that points due East. So which vector A becomes when I transport it to B depends on the path I take to transport it.

    This gives you the coordinate velocity of the particle. This coordinate velocity has no physical meaning. The coordinate velocity of light does not have to be ##c## in a curved spacetime (or even in a flat spacetime if you choose non-inertial coordinates). But that is not a matter of physics, it's a matter of your choice of coordinates.

    No, it isn't. It's the time coordinate in the coordinates you've chosen. It just happens that, for events on your worldline, the time coordinate is numerically equal to the time shown on your clock.

    We can. And the result we get will depend on which curve between points ##x## and ##x'## we choose to parallel transport ##V## along. The example I described above with a vector on the Earth's surface is using the process you describe, and illustrating how it is path dependent.
     
  16. May 24, 2017 #15

    Orodruin

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    There is no unique way of assigning "time" at any given event. You always have an ambiguity of which direction is "time".
     
  17. May 24, 2017 #16
    May we explore with a simple example?

    A beam of light is subjected to an equivalence principle scenario and its path is curved.
    Can we apply vectors prior to a coordinate system? If so, which vector represents the speed of light?

    1] - the vector "T" tangent to the curve? This vector "T" points "outside" the curve of the path ahead.

    2] - the vector "A" resulting from the addition of the component tangent vector and the component lateral vector? This lateral vector component of "A" points perpendicularly "inside" the curve of the path ahead, in the plane of the curve.

    Vector "T" has smaller magnitude than vector "A". At first it looks like whether observing at rest with respect to either a gravitating mass or an accelerating elevator, "T"<"A". But maybe upon relativity's imposition of coordinates (which happens when one adopts a reference frame?) both vectors "T" and "A" take magnitude c?
     
  18. May 24, 2017 #17

    PeterDonis

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    No, it isn't. It looks curved in space, but in spacetime its path is a geodesic, i.e., straight.

    Sure.

    Yes.

    No, it doesn't, because the light's path in spacetime isn't curved. See above.
     
  19. May 24, 2017 #18

    PeterDonis

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    [Note: edited to correct some items.]

    Aside from my previous comment, your example is looking at the wrong thing. The vector describing the "speed" of a light ray (or more precisely describing its "direction in spacetime" at a given event) is always the tangent vector to its worldline. (This is true of any object, not just light.) But what you want is a number describing, either (a) the measured speed of the light by an observer, or (b) the coordinate speed of light according to some coordinate chart.

    Here's how you get those numbers:

    (a) The observer must be co-located with the light ray at some event. At that event, you take the 4-velocity ##U## of the observer and form its inner product with the tangent vector ##T## of the light ray's worldline. This gives you a number, ##U \cdot T##. You then find a spacelike unit vector ##S## at the same event that is orthogonal to ##U## and points in the direction of the light ray, and find the inner product ##S \cdot T##. The ratio ##S \cdot T / U \cdot T## then gives you a number. That number will always be ##c## (or ##1## in natural units).

    (b) You must have a function ##\vec{x}(t)## that describes the motion of the light ray in the coordinate chart you are using, where ##t## is the time coordinate and ##\vec{x}## is a 3-vector giving the spatial coordinates. The coordinate speed of the light at a given event is then ##\vert d \vec{x} / dt \vert##, i.e., the magnitude of the vector you get by differentiating ##\vec{x})(t)## with respect to ##t##, evaluated at that event.
     
  20. May 24, 2017 #19
  21. May 24, 2017 #20

    PeterDonis

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    Note also that procedure (a) is equivalent to finding a local inertial coordinate chart centered on the chosen event, in which the observer with 4-velocity ##U## is momentarily at rest, and applying procedure (b) in that chart.
     
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