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Homework Help: Solution strategy for linearized gravity problem

  1. Apr 28, 2017 #1
    [Moderator's note: moved to homework forum.]

    Hello there,

    I am stuck at a problem, and I need some hits/solution strategy to get going. Suppose we consider linearized gravity, and there is some mass, ##M## at the origin at some unknown coordinates the metric is
    $$
    ds^2 = -(1 + 2 \phi)dt^2 + (1-2\phi)\delta_{ij}dx^i dx^j
    $$
    for ##\phi = -GM/(x^2 + y^2 + z^2) \ll 1##.

    Suppose that the mass ##M## now moves in the x direction with velocity ##v##.
    (1) What is the metric in this case
    (2) A photon is falling freely in the ##y## direction. I.e. its undeflected path is ## -b\vec e _x + t \vec e_y##. What angle is the actual photon trajectory deflected by?

    The first complication is that I dont know what the unknown coordinates are. How do I start this problem? The professor is giving a hint: for (2) Transforming to the rest frame of the moving mass and back is much easier than using the geodesic eqn.

    Kind regards,
    Marius
     
  2. jcsd
  3. Apr 28, 2017 #2

    Paul Colby

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    In linearized gravity I think the Lorentz transformation gives the answer.
     
  4. Apr 28, 2017 #3
    Can you be more specific? There are 3 sets of coordinates: 1 is the coordinates in which we know the placement of mass and the photon. Call these coordiantes ##x^\mu##. There is the frame of the stationary mass ##\xi^\mu## and the frame where the mass is moving, in which we somehow obtain the metric from problem (1), call these coordinates ##\zeta^\mu##. How do I find the coordinates of the photon in the ##\zeta^\mu## coordinates? It should be possible to boost these to the ##\xi^\mu## coordinates, then somehow work out the trajectory, and finally transform back to ##\zeta^\mu## and eventually to ## x^\mu##?
     
  5. Apr 28, 2017 #4

    Paul Colby

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    The answer to (1) is take your metric and do the transform? Let's pretend I know what the coordinates of a photon are, take them in one frame and transform them to another.
     
  6. Apr 28, 2017 #5

    PeterDonis

    Staff: Mentor

    No, it doesn't. The Lorentz transformation only works between two coordinate charts both of which are Minkowski, i.e., both of which have the metric ##\eta_{\mu \nu}##. That's certainly not true of the chart described in the OP, and I would not expect it to be true of the transformed chart that is being asked for either.

    Linearized gravity does not mean spacetime is approximated as flat. It only means that nonlinear effects of spacetime curvature are ignored (assumed to be too small to matter). But linear effects of spacetime curvature are still present.
     
  7. Apr 28, 2017 #6

    Paul Colby

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    true, what about the metric the OP has given?
     
  8. Apr 28, 2017 #7

    PeterDonis

    Staff: Mentor

    It's obviously not flat, because of the terms in ##\phi##. If you wanted to confirm that it's not flat, you could compute its Riemann tensor; it won't vanish.
     
  9. Apr 28, 2017 #8

    Paul Colby

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    Isn't ##\phi## specified to be small?
     
  10. Apr 28, 2017 #9

    Paul Colby

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    True, are you say linearized gravity doesn't hold ever? The form of ##\phi## given is both small and vanishes at spatial infinity. One may always change coordinates. Why will the Lorentz transform fail here?
     
  11. Apr 28, 2017 #10

    PeterDonis

    Staff: Mentor

    No, I'm saying that "linearized gravity" is not the same as "spacetime is flat". As I said in post #5.

    For a good quick treatment, see the beginning of Chapter 6 in Carroll's online lecture notes:

    https://arxiv.org/pdf/gr-qc/9712019.pdf

    Note carefully his expressions for the metric, Riemann tensor, and EFE in linearized gravity.

    Mathematically, of course you can apply any coordinate transformation you want to. Whether the transform is telling you anything meaningful physically is a different question. The latter is what I'm referring to when I say the LT does not "work" unless it's between global inertial charts in flat spacetime.
     
  12. Apr 29, 2017 #11
    So what do I do?
     
  13. Apr 29, 2017 #12

    Paul Colby

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    It's says in the problem statement you provided.

    My thought was change to the rest frame of ##M##. This is where the Lorentz transformation suggestion I was making which PeterDonis seemed to poo poo came from. That still seems reasonable to me given the rest of the problem statement but I wouldn't weight my opinion too much.

    Second thing of which I am more certain is to ask what path will the photon take and what equation would this path obey?
     
  14. Apr 29, 2017 #13
    We want to find the path based on the initial conditions given.

    Wikipedia says that Lorentz transformation is only correct for inertial coordinates. However, I was flipping though Gravitation by Misner et al.

    On page 439, it says that for linearized gravity

    $$
    \eta_{\alpha' \beta'} + h_{\alpha' \beta'}= g_{\alpha' \beta'} = \frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu}) = \eta_{\alpha' \beta'} + \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}h_{\mu \nu}
    $$
    If we can understand this, then it means we can transform the metric coefficients. Problem is, I dont understand that this equality is true:
    $$
    \frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu})
    $$

    Can you explain why it is?
     
  15. Apr 29, 2017 #14

    Paul Colby

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    Yes. In the linearized gravity where ##h_{\mu \nu}## is about a flat Minkowski metric ##\eta_{\mu \nu}## then ##h_{\mu \nu}## both transform as Lorentz tensors however the ##\eta_{\mu \nu}## is component wise Lorentz invariant.
     
  16. Apr 29, 2017 #15

    Paul Colby

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    Another question to ask is the bit about ##M## moving part of the problem you were given or did you add that bit as part of your struggle?
     
  17. Apr 29, 2017 #16
    I dont understand, how does this quote explain

    $$
    \frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu})?
    $$

    We dont know anything about the ##\frac{\partial x^\mu}{\partial x^{\alpha'}}##-factors. On the contrary, to me it sounds like your quote is a statement about lorentz invariance. I am confused.

    Thanks.
     
  18. Apr 29, 2017 #17

    Paul Colby

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    The ##\frac{\partial x^\mu}{\partial x^{\alpha'}} = \Lambda^\mu_{\alpha'}## are the Lorentz transformation. Also,

    ##\Lambda^\mu_\nu\Lambda^\alpha_\beta\eta_{\mu \alpha} = \eta_{\nu \beta}##​
     
  19. Apr 29, 2017 #18

    Paul Colby

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    Also, ask yourself, is the metric you are given one in which the mass is moving? Looks to me like ##M## hangs out at x=y=z=0.
     
  20. Apr 29, 2017 #19
    That is remarkable! Can you provide a reference/explain that it is true for the coordinates ##x^\mu## and ##x^{\mu'}##?

    According to the problem description, the metric is for a (stationary mass)/(rest frame of mass)
     
  21. Apr 29, 2017 #20

    Paul Colby

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    Most of this stuff really is. Read the section on linear gravity. MTW is quite good. Also there is a section devoted to Lorentz transformations as I recall.
     
  22. Apr 29, 2017 #21

    PeterDonis

    Staff: Mentor

    Having reviewed MTW, yes, I agree this is correct. However, if you look at Chapter 18, you will see an important caveat (which is what I think I was thinking of when I expressed doubt before about the Lorentz transformations being correct for this case): you have to choose the gauge conditions given by equation (18.8a) in MTW. This is a further restriction on the coordinates and the properties of the metric perturbation ##h_{\mu \nu}##, over and above the basic assumptions of linearized theory (ignoring quadratic and higher terms). MTW calls this further restriction the "Lorentz gauge" by analogy with electromagnetism; so the fully correct way to express what is said in the above quote is that ##h_{\mu \nu}## transforms like a Lorentz tensor in flat spacetime provided we are working in the Lorentz gauge.
     
  23. Apr 29, 2017 #22
    Alright, I take it then that

    $$
    \frac{\partial x^\mu}{\partial x^{\alpha'}}\frac{\partial x^\nu}{\partial x^{\beta'}}g_{\mu \nu} = \Lambda^\mu_{\alpha'}\Lambda^\nu_{\beta'}(\eta_{\mu\nu}+ h_{\mu\nu}) \tag{1}
    $$

    follows from using the Lorentz gauge then? Are we able to prove ##(1)##?
     
    Last edited: Apr 29, 2017
  24. Apr 29, 2017 #23

    Paul Colby

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    This is the standard general coordinate transformation specialized to the Lorentz case. It's more of a definition than a proof? Have you tried the second part of the problem. This would be finding the photon trajectory when the mass is stationary. You sound unsure about coordinate transforms and what they mean. You may need to back up a bit and work it through. Have a look at MTW section 2.9 page 66 and following. I bought this book back in the 70's and found it a steep learning curve. I should probably reread much of it.
     
  25. Apr 29, 2017 #24

    Paul Colby

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    This is a really important deal and I'm aware of it. Mr Jonsson need not be concerned because he's been given a geodesic problem which is solvable given a metric. I'm just trying to get him to say "geodesic" so he can get on track :wink: with his homework.
     
  26. Apr 30, 2017 #25
    Ok, I have been working away at this. I found the metric in the non moving frame. But how do I go from there. Suppose I write the up the geodesic equation or equations for conservation of momentum along a geodesic
    $$
    0 = g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda},
    $$
    We don't have an equation for ##\frac{dx^\mu}{d\lambda}##? How can we find this quantity in the unknown coordinates when they are unknown? Here is a drawing of the situation

    FullSizeRender.jpg
     
    Last edited: Apr 30, 2017
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