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How should I think about null geodesics?

  1. Jul 10, 2015 #1
    I am kinda being thrown into pretty intense physics and this really doesn't have too much to do with what I'm doing but I was wondering if null geodesics have zero length, what are the other dimensions or parameters that accounts for the apparent movement of particles? I am a visual learner and couldn't find any good visuals.
     
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  3. Jul 10, 2015 #2

    Nugatory

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    The only thing that moves on null geodesics is light - everything else moves on timelike and hence non-null worldlines (which are not necessarily geodesics, although they will be if the object is in free fall).

    It makes no sense to talk about the movement of particles without saying what that movement is relative to. That is, we cannot say "X is moving", we can only say "X and Y are moving relative to one another". You can try visualizing that by considering the worldlines that the two particles are moving along.
     
  4. Jul 10, 2015 #3
    So that's where length comes in and how it isn't existent.
     
  5. Jul 10, 2015 #4
    Basically bc of how u think about particles
     
  6. Jul 10, 2015 #5
    I just wanted some clarity. What is the physical reasoning for why particles move relative to one another and how is this different than lets say a planet and its sun? I thought we think about movement now as being an object moving relative to another. Is it because of the speed and the light properties of those particles?
     
  7. Jul 10, 2015 #6
    I think I got my questions mixed up. I guess I meant light and I just don't really understand the zero length idea then. Sorry guys!
     
  8. Jul 10, 2015 #7

    PeterDonis

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    Distance and time according to observers or instruments that measure the light. Remember that "length" in spacetime, in general, is the difference between the squared time interval and the squared space interval; more precisely, this is the "squared length" ##ds^2 = dt^2 - dx^2## (if we restrict to one space dimension). So all "zero length" (more precisely, zero squared length) really means is that the time interval equals the space interval--in other words, the distance that the light travels (when measured by some observer) is equal to the time it takes the light to travel that distance (when measure by the same observer). Note that this is in "natural units", where the speed of light is 1. This is ultimately a consequence of the fact that light travels at the same speed in all inertial frames.
     
  9. Jul 10, 2015 #8

    Nugatory

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    No difference at all (as long as we keep quantum mechanics out of the picture).
     
  10. Jul 11, 2015 #9
    Great thank you! I guess saying that statement is more conceptual which is what was tripping me up and you have to know the underlying concepts. For everything else, they are constrained by the time interval and the space interval to find the total length. I guess that is a geodesic in that case. I think I was thinking length as the length of a rectangle or the length of a football field. So anything else doesn't move as fast they do in time as they do to travel a distance basically.
     
    Last edited: Jul 11, 2015
  11. Jul 11, 2015 #10
    Please let me know what you think :)
     
  12. Jul 11, 2015 #11

    Mentz114

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    As PeterDonis said explicitly, there is a kind of length defined in SR that is not the usual spatial distance ##s=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}## but involves time as if it were a dimension. It is given by ##\tau=\sqrt{(t_1-t_2)^2 -(x_1-x_2)^2-(y_1-y_2)^2}##. This expression inside the radical is the difference between the squares of the time taken to move the spatial distance ##s## and ##s^2##. If the time taken is equal to the distance then ##\tau## is zero and the velocity ##v## is 1. For something with mass ##\tau^2>0##, and ##v<1## because the time taken is always greater than the distance travelled. So

    Light : time taken = distance travelled
    Masses : time taken > distance travelled.

    None of this has any connection with geodesics. That is a much more difficult idea.
     
    Last edited: Jul 11, 2015
  13. Jul 11, 2015 #12
    That's all I wanted to know :). Messed up w what I was trying to say. I thought Mass would make time taken greater as that makes sense. Thank you! This is so cool. Also, what Peter said was very explicit. I just wanted to ask a question about when mass is involved.
     
  14. Jul 12, 2015 #13

    vanhees71

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    It's just the usual sloppyness of physicists to call the fundamental form of the four-dimensional pseudo-Riemannian (the emphasis lies on pseudo!) a "metric", but it's not a true metric, because it's not positive definite but has 1 positive and 3 negative eigenvalues (or the other way around, but I'm a follower of the west-coast convention; for GR I use the conventions of Adler et al).

    This is of utmost importance since it provides the space-time with a causal structure. If there are no interactions (gravity in GR is not taken as an interaction but is discribed by the curvature of spacetime) massive particles move on geodesic time-like world lines.

    Light is special, because it's not describable as particles. In no sense of modern physics; photons are far from being particles in a classical sense and shouldn't be treated as such although that's done particularly in many textbooks on GR; I've just done some recitation sessions for the cosmology lecture. In preparing the problem sheets I was pretty surprised that I couldn't find a book where not the very misleading naive photon picture a la Einstein 1905 is used to derive important elements of the theory like the Hubble redshift and the luminosity distance and finally the luminosity-distance-redshift relations for a given cosmological model. All the results are, of course, correct, but the derivation is pretty misleading, providing a qualitative picture of light which is completely obsolete with the discovery of modern relativistic quantum field theory in the late 1920ies by Dirac, Heisenberg, Pauli, Born, Jordan et al.

    The fully adequate treatment of light in the context of GR (bending of light rays on the sun, gravitational redshift on earth, Hubble redshift in cosmology, etc.) is, however, classical electrodynamics in curved-spacetime. Here, fortunately it's sufficient to use the eikonal approximation, which finally is equivalent to the naive Einstein-photon picture due to a mathematical analogy. The eikonal approximation is valid, if the wave length of the considered electromagnetic wave is small compared to the typical length scales of the physical situation. Here this is the typical length scale along which the index of refraction changes and the curvature scalar (Ricci scalar) of the metric. One shouldn't also look to close to opaque objects, because there always wave optics is necessary to describe the detailes of the defraction pattern.

    The eikonal approximation describes the phase ##\psi## of the electromagnetic wave by the partial differential equation
    $$g^{\mu \nu} (\partial_{\mu} \psi) (\partial_{\mu} \psi)=0.$$
    This can be read as the Hamilton-Jacobi equation for a massless particle, i.e., the "naive photons" a la Einstein. In terms of classical optics these are the rays of ray optics. The space-time dependent wave four-vector is given by
    $$k_{\mu}=\partial_{\mu} \psi,$$
    and the light rays by the equation
    $$\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda}=k^{\mu},$$
    where ##\lambda## is an appropriately scaled world-line parameter.

    From the eikonal equation it follows that
    $$\frac{\mathrm{D} k^{\mu}}{\mathrm{D} \lambda}=\frac{\mathrm{D}^2 x^{\mu}}{\mathrm{D} \lambda^2}=\frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \lambda^2} + \Gamma^{\mu}_{\rho \sigma} \frac{\mathrm{d} x^{\rho}}{\mathrm{d} \lambda} \frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \lambda}=0,$$
    which means that the light rays are given by null geodesics.
     
  15. Jul 12, 2015 #14
    A particle moves relative to another particle or a planet like earth moves relative to our sun because either distance or position changes.

    One important component is whether spacetime is flat, without any gravity, or whether it is 'curved', as when the is gravity present. Einstein taught us spacetime is in general 'curved' and it has been found there are many different types of curvatures causing geodesics to vary to from straight line. Without any gravity, a particle, or a planet, in motion remains moving along a straight line; If gravity is present, we use the term 'geodesic' [ a generalization of the idea of a straight line path to a curved path] to describe the motion in curved spacetime. It turns out that Einstein also figured out an 'equivalence' principle between gravity and acceleration, another component of motion.

    We say the earth revolving about the sun is accelerating because it's direction constant changes....to maintain a basically circular orbit. But at the same time both the sun and earth are also moving respect to the moon, for example, and our entire galaxy.

    A neat visual way to picture motion with and without gravity has been explained in these forums: Acceleration curves spacetime in such a way that it can be represented as a curve on flat graph paper. Gravity curves spacetime in a more complex manner in such a way that the graph paper itself must be curved...say via hills and valleys....it can no longer remain flat when visualizing gravitationally curved spacetime and motion. .
     
  16. Jul 12, 2015 #15

    PeterDonis

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    Only in a coordinate sense. The Earth is in free fall in its orbit about the sun; it feels no acceleration. Acceleration that is felt, i.e., proper acceleration, is the only kind of acceleration that is invariant; coordinate acceleration is not, it depends (as the name implies) on the coordinates you adopt.

    No, it doesn't. Proper acceleration curves the worldline of whatever is experiencing it. But coordinate acceleration does not curve anything. The Earth, in its orbit around the sun, is following a straight path in spacetime.
     
  17. Jul 12, 2015 #16

    pervect

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    I'm not aware of any good visuals, but I have a few comments.

    Let's consider initially a 2d flat (Minkowskii) space-time with one spatial dimension (x), and one time dimension (t). As an alternative to specifying the position of an object with (x,t) we can specify the values u = (x-ct) and v = (x+ct).

    This isn't "visual", but hopefully it's simple enough mathematically not to be a real shocker. We can write that x = (u+v)/2 and t = (v-u)/2c, it's just linear algebra.

    Now for a physical interpretation of this that also (hopefully) ties this seemingly unrelated observation to your question. On any null geodesic progressing in the +x direction with time, we can write (x-ct) = u = constant. We can thus regard u as a parameter, one that picks out a specific null geodesic, a parameter which is constant everywhere along a geodesic propagating in the +x direction, on that selects a particular null geodesic. We call this parameter, u, a null coordinate. Similarly, for geodesics moving in the other direction, the -x direction, (x+ct) = v is constant, and v is a second null coordinate.

    So as an alternative to using space and time (t,x) as coordinates, we can use a pair of null coordinate (u,v).

    If we want to extend this to 4 dimensions (t,x,y,z), the easiest approach is to use (u,v) as null coordinates and keep (y,z) as spatial coordinates.

    If you really, really want to make all coordinates null, there's a sort of hack that lets you do this. It turns out to be a useful hack, but it requires complex numbers, making it rather abstract, even more abstract than the above approach. For more details see https://en.wikipedia.org/wiki/Construction_of_a_complex_null_tetrad
     
  18. Jul 12, 2015 #17
    Hey, Peter, quit it: How did I get picked on but not Nugatory in his post #8?? [LOL]

    I hope not but I suspect in aggregate we have all pretty well confused Wonderer.

    For the benefit of the Wonderer, am I correct that by 'straight path' you mean earth orbit around the sun follows a geodesic in curved space-time......??

    That seems a better way to answer to Wonderer's questions in posts #1 and 5 than what I posted previously.

    Good insight, thanks.
     
  19. Jul 12, 2015 #18

    PeterDonis

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    Yes.
     
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