Relativity 109d: Gravitational Waves - Transverse-Traceless Gauge (Plus and Cross Polarizations)

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  • เผยแพร่เมื่อ 26 พ.ย. 2024

ความคิดเห็น • 34

  • @StefSubZero270
    @StefSubZero270 2 ปีที่แล้ว +6

    Very good video, the first time i studied this part I felt like i did not understand very well some parts but this cleared up my thoughts. Cheers!

    • @eigenchris
      @eigenchris  2 ปีที่แล้ว +3

      Thanks. This one took around 10 days longer than I wanted because I was double and triple checking things.

  • @sonarbangla8711
    @sonarbangla8711 2 ปีที่แล้ว +1

    In other words gravitational wave spacing is CONSTANT and non diminating, like all other waves, so they pervade the whole of space up to infinity, without weakening. Very well explained, thank you.

    • @eigenchris
      @eigenchris  2 ปีที่แล้ว +2

      I think real gravitational waves do weaken as they travel. I just discussed plane waves in this video because they are the simplest theoretical example.

    • @sonarbangla8711
      @sonarbangla8711 2 ปีที่แล้ว +1

      @@eigenchris Electromagnetic waves from distant stars are visible, no matter how far. Only those falling outside Penrose cone will become invisible to us, I don't think they weakens, maybe because they are both bosons.

    • @Astro-X
      @Astro-X 2 ปีที่แล้ว

      Great question! For an electromagnetic wave it is tightly confined in a small region that is possessed by its corresponding photon area. There is no loss in energy (in a vacuum) so the energy is conserved (the frequency may change due to Doppler shift though).
      However for a gravitational wave, I imagine that the wave is not confined to a small region and pervades the whole spacetime. The initial energy that was released into the gravitational wave gets dilated very very quickly as the distance it travels through increases and the actual intensity we measure is small.
      The spacing of the waves is constant in comoving coordinates however when it comes across a star or other massive object there may be some interference pattern from that location. Plus if we're talking about the expansion of the universe then the spacing gets stretched out as time goes on.

  • @mrmilionphysics8342
    @mrmilionphysics8342 2 ปีที่แล้ว +5

    Great!!
    Keep going king ❤️🔥

  • @jeancorriveau8686
    @jeancorriveau8686 6 หลายเดือนก่อน

    I've been scratching my head to understand while reading other sources with little success. This video makes me grasp the context. As for the GR equations, maybe eigenchris provides other videos.

  • @imaginingPhysics
    @imaginingPhysics 2 ปีที่แล้ว +2

    In what sense do gravitational waves carry energy? I have understood that "gravitational energy" is not a well defined thing in GR. The energy momentum tensor does not take it into account for example. And by a coordinate transformation one can shift into a frame where there is no "gravity", hence no gravitational energy.
    Yet it has some kind of energy, since a black hole binary "loses energy" in gravitational waves which causes them to collide.
    So, what energy do g-waves carry?

    • @dominicellis1867
      @dominicellis1867 2 ปีที่แล้ว +3

      Momentum in a gravitational wave is propagated as space itself so the energy is just the potential difference between the peak and trough of the wave through minkoswki space time. That being said, without a formalized definition of the fundamental particle of gravity it’s hard to say what exactly is waving on the quantum level assuming that unlike light and the nuclear forces, space time either changes or it doesn’t there’s no superposition state of the minkoswki metric tensor

  • @ChronusZed
    @ChronusZed ปีที่แล้ว +1

    Am I correct in understanding the geometric idea behind the TT gauge (compared to a generic Lorenz gauge) is that we're essentially forcing (1) translations along U should generate geodesics, and (2) the metric volume should match the Minkowski-Euclidean volume to first order?

    • @Temperancefp
      @Temperancefp 3 หลายเดือนก่อน

      This is a very interesting question

  • @TheJara123
    @TheJara123 2 ปีที่แล้ว +2

    As always brilliant man...,,,👍

  • @Schraiber
    @Schraiber 2 ปีที่แล้ว +2

    Great video as always! I guess a couple things aren't obvious to me:
    1) How do we know the transverse-traceless gauge is a Lorenz gauge? I suppose one can just crank through some calculus and algebra from the definitions and it works?
    2) In practice, we choose a coordinate system to measure gravitational waves experimentally (like at LIGO). Is the choice we make a Lorenz gauge?

    • @eigenchris
      @eigenchris  2 ปีที่แล้ว +5

      1) The TT Gauge requires the Lorenz Gauge condition to be true. So you start with the Lorenz Gauge (4 constraints) and then add the additional 4 TT Gauge constraints.The TT Gauge is just a Lorenz Gauge coordinate system with 4 additional constraints on it.
      2) My ability to answer LIGO questions is a bit limited, but I'll do my best based on what I currently know. I mentioned gravitational waves have 2 polarizations. Based on my work in the next video (109e), a given LIGO detector will be "blind" to one of these polarizations. For example, if it sees the + polarization, it will be unable to see the x polarization. The LIGO detectors ultimately detect photons. The number of photons they see is correlated with the "stretching" and "squashing" of spacetime that the + polarized causes. So as far as I know, all they're really doing is watching a photon count going up and down overtime, and inferring the existence of gravitational waves from this. I don't think any particular spacetime coordinate system would need to come into play here. All the theory in my videos is necessary to propose the existence of gravitational waves and know the waveforms we should expect to see, but I don't know how much of it actually comes into play for the actual experimental measurements. I can't promise anything I said above is true, but it's my best guess. You'd probably need to find an actual physicist who works with GR waves to get a proper answer.

    • @Schraiber
      @Schraiber 2 ปีที่แล้ว

      @@eigenchris thanks that super useful and helpful!

  • @utof
    @utof 2 ปีที่แล้ว +7

    One day i will get to this... still didnt get tensors tho :D

    • @Astro-X
      @Astro-X 2 ปีที่แล้ว

      You must watch Keenan Crane Discrete Differential Geometry to better understand it

    • @utof
      @utof 2 ปีที่แล้ว

      ​@@Astro-X woah i forgot about this comment. Thanks for the recommendation tho! i did stumble upon it eventually and i think i get tensors now :D (i even made a video about one physics theory to really test my knowledge heh)

  • @mongooseman2319
    @mongooseman2319 2 ปีที่แล้ว +2

    9:00 : "since gravitational waves are waves in the metric, and the metric is symmetric (...)" hahaha

  • @keshavshrestha1688
    @keshavshrestha1688 ปีที่แล้ว

    Great video, After 19:29, How we can be sure that two independent components are non zero components of polarisation tensor? Is it not necessary to check whether two independent components are zero or non zero?

    • @eigenchris
      @eigenchris  ปีที่แล้ว

      They can be zero in some locations, just as the electric magnetic field can be zero in some locations. But if some mass wobbles back and forth very quickly (as with two black holes orbiting each other and about to merge), gravitational waves can be created. Same goes for vibrating electric charge and electromagnetic waves.

    • @keshavshrestha1688
      @keshavshrestha1688 ปีที่แล้ว

      @@eigenchris sorry I didn't get you, in this lecture video at last we get A_xx and A_xy as a two independent components of polarisation tensor but there is a difference in being independent and being non-zero, well you have showed that these two components are independent but you haven't showed that these two independent components are also non-zero so my question is, is it not necessary to show whether these components are non-zero or zero? Or using all constrain equation due to degrees of freedom we let to the conclusion that these two independent components are now non-zero, is that the reason that two independent components are non-zero?

  • @darshkhandelwal
    @darshkhandelwal 2 หลายเดือนก่อน

    Why do we need the orthogonality condition? Can't the angle of the velocity of the observer and the gravitational waves be different than this condition? Pls help

    • @eigenchris
      @eigenchris  2 หลายเดือนก่อน

      Yes, but the equations are simplest when the frame is orthogonal to the wave. You can always change frame to a different observer if you want.

  • @cirno6358
    @cirno6358 ปีที่แล้ว

    I'm confused about 9:43, shouldn't that expression mean that the ROWS of the amplitude matrix should be orthogonal to the wave covector components?

    • @cirno6358
      @cirno6358 ปีที่แล้ว

      and at 12:34, I don't think you can express it with traditional matrix multiplication, as it involves summing the different components of each column of A with the components of U, whereas matrix multiplication like that involves summing the components of each row of A with the components of U.

    • @cirno6358
      @cirno6358 ปีที่แล้ว

      oh wait, is it because A is symmetrical so it doesn't matter whether you do its rows or columns?

  • @longsarith8106
    @longsarith8106 2 ปีที่แล้ว

    Excuse me , Sir! I don't follow at point 14:31. Why taking trace on amplitude (A) gives us zero?

    • @eigenchris
      @eigenchris  2 ปีที่แล้ว

      At around 10:50 I say that the Lorenz Gauge is a class of coordinate systems that comes with 4 degrees of freedom. We take 3 of these to be "transverse" constraints and we decide the last constraint will be to make A have trace zero.

  • @jigold22571
    @jigold22571 2 ปีที่แล้ว

    Thank You 🕊

  • @pocaudraphael6066
    @pocaudraphael6066 2 ปีที่แล้ว

    👍

  • @vinnyhorapeti2461
    @vinnyhorapeti2461 2 ปีที่แล้ว +1

    Sir please watch gary yourofsky's speech on veganism on TH-cam and end animal cruelty also check out earthling Ed, Dr michel greger and dominion documentary 🙏🏻 💗 💚