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Einstein's 1911 prediction (Gravitational Lensing)

  1. Jun 30, 2011 #1


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    Hi. I am trying to make my way through


    one step at a time if I have too.

    I already got stuck once, where it says:

    The paper makes it sound like these two variables are constant, but the way it should be defined is

    "Let t1 be the time at the bottom of the elevator. Let t2 be the time at which a photon emitted from the bottom of the elevator at time t1 arrives at the top of the elevator."

    That way neither of the variables is a constant, and it makes sense to do impliciti differentiation.

    Now I am stuck at another step, where it says:

    When I solved equation (1) for t2 when t1=0, I got:

    [tex]t_2 = \frac{2 c}{a} \pm \frac{c}{a}\sqrt{\left ( 2 - \frac{4 a L}{c^2} \right )}[/tex]

    and I can't see how


    evaluates to

    [tex]\frac{1}{\sqrt{1-\frac{2 a L}{c^2}}}[/tex]
    Last edited: Jun 30, 2011
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  3. Jun 30, 2011 #2


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    I am thinking there may be some kind of Taylor expansion going on, or something? Or am I just missing some simple algebraic identity?
  4. Jun 30, 2011 #3

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    Hi JDoolin! :smile:

    You didn't solve equation (1) properly.
    If you do, you'll find everything follows by simple algebraic manipulation.

    The proper solution is:
    [tex]t_2 = \frac c a (1 \pm \sqrt{1- \frac {2 a L} {c^2}})[/tex]
  5. Jun 30, 2011 #4


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    Looks to me like your solution for t2 is off. Right hand side needs to be divided by 2. Then, with a little rearrangement, you get:

    t1 = (c/a) - sqrt((c/a)^2 - 2L/a)

    This, then leads to:

    1 - (a/c)t2 = sqrt( 1 - 2aL/c^2)

    Then, note that when substituting in dt2/dt1, they say this should also be evaluated at t1=0, which immediately makes the numerator 1; they the above leads to what they say.
  6. Jun 30, 2011 #5


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    Oh, my. I made a newbie mistake (four times in a row.)

    I used [tex]x = -b \pm \sqrt{\frac{b^2 - 4 a}{2a}}[/tex]

    instead of the correct equation

    [tex]x = \frac{-b \pm \sqrt{b^2 -4 a c}}{2 a}[/tex]
  7. Jun 30, 2011 #6


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    ... and then you can use the first order Taylor expansion:

    [tex]x = \frac{2 a L}{c^2}[/tex]

    [tex]f(x)=\sqrt{1-x} [/tex]

    [tex]f(x)\approx f(0)+x f'(0) = 1 - \frac{a L}{c^2}[/tex]
  8. Jul 1, 2011 #7


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    Now, much of the remaining argument deals with Huygen's Principle. I am attaching an image which (I think) gets across the gist of this idea. The bottom edge of a beam of light will be a slower speed of light and a faster frequency.

    The top edge has a faster speed of light, and a lower frequency.

    The two combine to turn the wave-front toward the mass, and the complicated part that is still to come is to calculate exactly how much.

    Attached Files:

  9. Jul 1, 2011 #8

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    Do you have a question?
  10. Jul 1, 2011 #9


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    I am wondering how
    [tex]1 - \frac{a L}{c^2}[/tex]

    evaluates to:

    [tex]1 - \frac{m}{r^2}[/tex]

    [STRIKE]and what kind of conversion factor you would use to change G in mks units to G in c=1,G=1 unts.[/STRIKE] (Actualy that's a silly question; you'd just take the reciprocal.)

    After that, I think I can change a product of ratios into an integral (by taking the logarithm), do a 1st order taylor series expansion of the integrand, integrate, and change back to a ratio (by taking e^answer), do a 1st order taylor series expansion again and come up with the ratio

    [tex]\frac{\nu_2}{\nu_1}=1 + m\left ( \frac{1}{r_2}-\frac{1}{r_1} \right )[/tex]
    Last edited: Jul 1, 2011
  11. Jul 1, 2011 #10


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    So 1 - (a L)/c^2

    replace a=m/r^2

    replace c=1

    replace L=1?

    I guess that would be obvious, but the trouble is, I can't quite wrap my head around it. It's like we're replacing one big elevator of height L, and acceleration a, with a bunch of differential elevators with height dr, and acceleration m/r^2. Does that make any sense?
  12. Jul 1, 2011 #11

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    Basically you seem to have puzzled everything out already. :)

    Where did you get dr from?

    As I see it, they just took an elevator at some distance r of some large mass m.
    Like an elevator somewhere above earth's surface.
  13. Jul 2, 2011 #12


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    But this "elevator" is huge. We have an "elevator" that extends from distance r1 from some large mass m, to a distance r2 from some large mass m.

    Since the gravitational strength varies between point r1 and r2, we can't just use a single value for a. We have to treat a as a function of r, and do an integral to find the combined effect.
  14. Jul 2, 2011 #13


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    I'm still a little ways away from puzzling it out, though.

    The article derives the equation:

    [tex]\frac{\nu_2}{\nu_1}= 1 + m \left ( \frac{1}{r_2}-\frac{1}{r_1} \right )[/tex]

    Then the http://www.mathpages.com/rr/s8-09/8-09.htm" [Broken] says
    I can work on the probem some more (trying out Huygen's Principle (See post #7) and seeing what I can come up with), but I don't see any information in the article about how Einstein came to the conclusion of θ=2m/R. (Whoops! hold on. I think I missed something like the WHOLE REST OF THE ARTICLE.) The foray into Tensor notation confused me into thinking he was done.
    Last edited by a moderator: May 5, 2017
  15. Jul 2, 2011 #14


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    I have to put it in my own words to understand it.

    [tex]\frac{\nu_2}{\nu_1}= 1 + m \left ( \frac{1}{r_2}-\frac{1}{r_1} \right )[/tex]

    Instead of talking about the gtt element of the metric tensor, I would rather use a straightforward approach.

    [tex]\frac{\nu_{\infty}}{\nu_1}= 1 + m \left ( \frac{1}{\infty}-\frac{1}{r_1} \right ) = 1 - \frac{m}{r_1} = \frac{1/dt}{1/d\tau_r } = \frac{d\tau_r }{dt}[/tex]

    The idea here is that the speed of time in a gravitational well, is based relative to a universal speed of time, somewhere far, far away from any gravitating bodies. So the speed of time near (a distance r) from a gravitating body, can be expressed as a ratio with the universal speed of time (a distance infinity) from the gravitating body dt.

    [tex]\frac{d\tau }{dt} \approx\sqrt{1-\frac{2 m}{r}} \approx 1 - \frac{m}{r}[/tex]

    And here's where Einstein's calculation was wrong:

    [tex](d\tau)^2=\left (\frac{\partial \tau}{\partial t} \right )^2 dt^2 - dx^2 - dy^2 - dz^2[/tex]

    That's usually described as a problem with the special theory of relativity, but I wonder whether it isn't just a partial differential equations error?

    Wouldn't the full expression be:

    [tex](d\tau)^2=\left (\frac{\partial \tau}{\partial t} \right )^2 dt^2 - \left(\frac{\partial \tau}{\partial x} \right )^2 dx^2 - \left(\frac{\partial \tau}{\partial y} \right )^2 dy^2 - \left(\frac{\partial \tau}{\partial z} \right )^2dz^2[/tex]

    ... regardless of whether you were using Special Relativity or General Relativity?
  16. Jul 2, 2011 #15


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    It leads me to another part of the paper


    Section 6-6

    Where it appears that there are terms involving

    [tex]\frac{\partial^2 \tau}{\partial x \partial y} dx dy, \frac{\partial^2 \tau}{\partial x \partial z} dx dz, \frac{\partial^2 \tau}{\partial y \partial z} dy dz[/tex]

    I'm not sure whether we're using

    [tex]\left (\frac{\partial \tau}{\partial x} \right )^2, \left (\frac{\partial \tau}{\partial y} \right )^2, \left (\frac{\partial \tau}{\partial z} \right )^2[/tex]


    [tex]\frac{\partial^2 \tau}{\partial x^2},\frac{\partial^2 \tau}{\partial y^2},\frac{\partial^2 \tau}{\partial z^2}[/tex]

    ...probably the latter.

    That would make he last equation in my last post something more like:

    [tex](d\tau)^2= \frac{\partial^2 \tau}{\partial t^2} dt^2 - \frac{\partial^2 \tau}{\partial x^2} dx^2 - \frac{\partial^2 \tau}{\partial y^2} dy^2 - \frac{\partial^2 \tau}{\partial z^2} dz^2 + \text{other terms}[/tex]
    Last edited: Jul 2, 2011
  17. Jul 3, 2011 #16


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    I'm taking a detour through the appendix: http://www.mathpages.com/rr/appendix/appendix.htm" [Broken].

    I thought part 2 was interesting, and I thought maybe I had a question, but as so often happens, by the time you formulate a question carefully, the answer comes to you.

    When you take a function in two variables, z=f(x,y) and take the total derivative, you get:

    [tex]dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial x} dz[/tex]

    If [tex]f(x,y)=A x^m y^n[/tex]


    [tex]\frac{\mathrm{d} z}{\mathrm{d} t} = A x^m n y^{n-1}\frac{\mathrm{d} y}{\mathrm{d} t}+ A y^n m x^{m-1}\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]

    In general, though z=f(x,y) would not be of the form z=A xm yn

    But I guess it CAN be expanded using some kind of two-variable Taylor series into a polynomial expression,

    [tex]z = f(x,y) = \sum_{m,n} \left ( A_{m,n}x^m y^n\right )[/tex]

    and then something like the equation in the appendix section 2 can be used.

    Better yet, the equation could be expressed

    [tex]\frac{\mathrm{d} z}{\mathrm{d} t} = \frac{\mathrm{d} }{\mathrm{d} t}\sum_{m,n} \left ( A_{m,n}x^m y^n\right ) = \sum_{m,n} \left [A_{m,n}\left ( x^m n y^{n-1}\frac{\mathrm{d} y}{\mathrm{d} t}+ y^n m x^{m-1}\frac{\mathrm{d} x}{\mathrm{d} t} \right ) \right ][/tex]

    I feel like I've gone far astray from my main question, and perhaps I haven't really asked it clearly yet. My main question whether we must accept the premise that "space stretches" in the presence of gravitational fields, or if we can account for Einstein's 1911 error entirely in terms of "changes in the rate of time" in the presence of gravitational fields?

    Since Einstein did leave out such partials as ∂τ/∂x, ∂τ/∂y, ∂τ/∂z, it seems to me that these are changes in the rate of time; not changes in the scale of space. This suggests that we have no evidence, at least from this experiment, for any spatial scaling. But, of course, I may be missing something, and that is why I feel I have to understand this particular problem, step-by-step.
    Last edited by a moderator: May 5, 2017
  18. Jul 20, 2011 #17


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    Alright, back from my detour, at last; I think it helped somewhat. I can break the problem down into parts.

    The Schwarzschild metric in spherical coordinates:

    [tex]\partial \tau^2 = \left(1-\frac{2 m}{r}\right) dt^2 - \left( \frac{2m}{r-2m}\right) dr^2 -r^2 d\theta^2 - r^2 sin(\theta)d\phi^2[/tex]

    (1) I need to explain the coefficients [itex]\left(1-\frac{2 m}{r}\right)[/itex] and [itex]\left( \frac{2m}{r-2m}\right)[/itex] I've already got somewhat of a handle on the first coefficient (see post 12 and 14 in this thread, except I appear to have an extra 2.)

    (2) I need to express this metric in cartesian or polar coordinates, so that the path (r(t),Φ(t)) or(x(t),y(t)) of a particle can be analyzed.

    (3) To find the path of a photon, merely set dτ=0. Solving the differential equation will yield some relationship between (x,y,t) or (r,Φ,t) and in that relationship, I suspect, we'll find the path expected by general relativity.
  19. Jul 22, 2011 #18


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    Part 1a. Finding [itex]\left(1-\frac{2 m}{r}\right)[/itex]

    I have once again worked my way through http://www.mathpages.com/rr/s8-09/8-09.htm and want to document the tricky parts.

    The positions of the bottom and top of an elevator are described by: [itex]x_1 = \frac{1}{2} a t^2[/itex] and [itex]x_2 = L +\frac{1}{2} a t^2[/itex]

    It should first be noted that this is an approximation for low acceleration. If you want to treat for large accelerations, you need to make a hyperbola instead of a parabola:


    This is only the first of many, many, many approximations. But the rest of the approximations will be based on the Maclaurin series expansions:

    [tex]\begin{align} \displaystyle \mathrm{ Maclaurin\: Series\: Expansions} \\f(x) &\approx \frac{f(0) x^0}{0!} + \frac{f'(0) x^1}{1!} + \frac{f''(0) x^2}{2!} + ... \\\sqrt{1+x} &\approx \frac{1}{0!}+\frac{x/2}{1!}-\frac{x^2/4}{2!}+\frac{3 x^3/8}{3!}+... \\e^x &\approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}+... \end{align}[/tex]

    We imagine a photon of light leaving the bottom of the elevator at the moment when the elevator slows to a stop, before accelerating upward again. When does the path of the photon,

    [tex]x = c t[/tex]

    meet the path of the top of the elevator?

    [tex]x = L + \frac{1}{2}a t^2[/tex]

    Setting the two equal and solving for t, we find there are two solutions.

    [tex]t=\frac{c}{a}\left ( 1 \pm \sqrt{1-\frac{2 a L}{c^2}} \right )[/tex]

    The lower value of t is the one we want. The higher term is nonsense, because it represents the second time the elevator meets the photon. This would not happen if we used the hyperbolic equation for the position of the elevator, because the hyperbolic equation never reaches and passes the speed of light. The quadratic equation, however, lets the elevator exceed the speed of light, and run into the photon a second time, producing the nonsense solution.

    Our interest now turns to the question of how fast a clock at the top of the elevator runs compared to how fast a clock at the bottom of the elevator. This requires us to define two time variables. I don't know how standard this is, but I'd like to call them [itex]t_1(n)[/itex] and [itex]t_2(n)[/itex], being the time that the nth photon leaves the bottom of the elevator, and the time the nth photon arrives at the top of the elevator.

    I am noting this n variable because I think if we are to take a derivative of t1 and t2, we should keep in mind that there must be some dependent variable with respect to which t1 and t2 are changing.

    We note that

    [tex]c = \frac{\Delta x}{\Delta t} = \frac{x_2(t_2)-x_1(t_1)}{t_2 -t_1} = \frac{L+\frac{a}{2}(t_2^2-t_1^2)}{t_2 - t_1}[/tex]

    That is, the distance is the position of the top of the elevator at time 2 minus the position of the bottom of the elevator at time 1.

    [tex]\begin{align*} c ({t_2 - t_1})&= L+\frac{a}{2}(t_2^2-t_1^2)\\ c (dt_2 -dt_1) &= a (t_2 dt_2 -t_1 dt_1)\\ c dt_2 -a t_2 dt_2 &= c dt_1 -a t_1 dt_1 \\ (c -a t_2) dt_2 &= (c -a t_1) dt_1 \\ \frac{dt_2}{dt_1} &=\frac{c - a t_1}{c- a t_2}\\ \frac{dt_2}{dt_1} &=\frac{1 - a t_1/c}{1- a t_2/c} \end{align*}[/tex]

    However, what we are modeling is a gravitational field, and the idea of the equivalence principle is that if you're watching a descending elevator at the very instant, t1=0, that it ceases to decelerate, stops, and begins to accelerate; at the moment when both the top and bottom of the elevator simultaneously stop, this is a perfect model for watching what is happening to someone standing in a gravitational field.

    So, mathematically, what this means is we are going to set t1 = 0.

    [tex]\left ({\frac{d t_2}{d t_1}} \right ) _{t_1=0}= \left (\frac{1-\frac{a}{c} t_1}{1-\frac{a}{c} t_2} \right ) _{t_1 =0} =\frac{1}{1-\left ( 1 - \sqrt{1-\frac{2 a L}{c^2}} \right )} = \frac{1}{\sqrt{1-\frac{2 a L}{c^2}}}[/tex]

    Now think a moment about waves, with their frequencies, and wavelengths, and periods. What do the dt's represent? They are proportional to the periods of waves, and reciprocal to the frequencies of waves. So note that

    [tex]\frac{\nu_2}{\nu_1} = \frac{\mathrm{d} t_1}{\mathrm{d} t_2}=\sqrt{1 - \frac{2 a L}{c^2}} \approx 1 - \frac{a L}{c^2}[/tex]

    We don't want to assume, however, that the gravitational field a, within the elevator will be uniform. So what is done next is to divide the elevator into little chunks, noting, first of all, that within each chunk, [tex]a(r)=\frac{m}{r^2}[/tex].

    And in total, the value is a product:

    [tex]\frac{\nu_{top}}{\nu_{bottom}} = \prod_{bottom}^{top}\frac{ \nu_{upper} }{\nu_{lower}}[/tex]

    Now, did you know that the logarithm of the product of several terms is equal to the sum of the logarithms of those same terms?

    [tex]\begin{align*} \ln \left (\frac{\nu_{top}}{\nu_{bottom}} \right ) &=\ln \left (\prod_{bottom}^{top}\frac{ \nu_{upper} }{\nu_{lower}} \right ) \\ &= \sum \ln \left (\frac{ \nu_{upper} }{\nu_{lower}} \right )\\ &= \sum _a \ln(1-\frac{a L}{c^2}) \\ &=\sum _r \ln(1-\frac{m L}{r^2 c^2}) \end{align*}[/tex]

    Further, we can break this sum down continuously (and make our second Maclaurin Series Expansion) until we have an integral:

    [tex]\begin{align*} \sum _r \ln(1-\frac{m L_r}{r^2 c^2}) &=\int _{r_1}^{r_2} \ln \left( 1-\frac{m dr}{r^2 c^2} \right) \\ &\approx \int _{r_1}^{r_2} \left (-\frac{m d r }{r^2 c^2} \right )\\ &=-\frac{m}{c^2}\int _{r_1}^{r_2} r^{-2} dr\\ \ln \left ( \frac{\nu_2}{\nu_1}\right ) &=\frac{m}{c^2}\left ( \frac{1}{r_2}- \frac{1}{r_1} \right ) \end{align*}[/tex]

    Taking e to the power of both sides (and using a Taylor Series expansion of [itex]ln(1+x)=x[/itex]

    [tex]\begin{align*} \left ( \frac{\nu_2}{\nu_1}\right )&= Exp \left ({\frac{m}{c^2}\left ( \frac{1}{r_2}- \frac{1}{r_1} \right )} \right )\\ &\approx 1 + \left ({\frac{m}{c^2}\left ( \frac{1}{r_2}- \frac{1}{r_1} \right )} \right ) \end{align*}[/tex]

    Now, defining dt as the period of time far away from the mass, and d\tau as the period of time at distance r from the mass:

    [tex]\begin{align*} \left ( \frac{d \tau}{d t}\right )= \left ( \frac{\nu_\infty}{\nu_r}\right ) &= Exp \left ({\frac{m}{c^2}\left ( \frac{1}{\infty}- \frac{1}{r} \right )} \right )\\ &\approx 1 - {\frac{m}{c^2 r}} \end{align*}[/tex]

    And we are interested in finding the square of this quantity:

    [tex]\begin{align*} \left ( \frac{d \tau}{d t}\right )^2 &\approx \left (1 - {\frac{m}{c^2 r}} \right )^2\\&\approx 1 - \frac{2 m}{c^2 r} \end{align*}[/tex]

    Set c=1, and we have the value given by the Schwarzschild metric:

    [tex]g_{11} = \left ( \frac{d \tau}{d t}\right )^2 = 1 - \frac{2 m}{ r}[/tex]

    If I counted right, there are six approximations. The use of a parabola instead of a hyperbola. Four Maclaurin Series approximations, and one Taylor series approximations.
    Last edited: Jul 22, 2011
  20. Jul 23, 2011 #19


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    Continuing through the article, http://www.mathpages.com/rr/s8-09/8-09.htm ,
    I gather that Huygen's Principle can be written in the form:

    [tex]\mathrm{Total \: Deflection} = \int_{-\infty}^{\infty} \frac{\partial c}{\partial y} dx[/tex]

    Though I'm not precisely certain whether the result is meant to be a change in angle or a change in distance, it should relate somehow to the primitive drawing I made:


    To find ∂c/∂y, we take

    [tex]d\tau^2 = \sqrt{\left ( 1-\frac{m}{r} \right )^2}dt^2 - dx^2 -dy^2[/tex]

    and set [itex]d\tau=0[/itex] for the case of the speed of light so:

    [tex]dx^2 + dy^2 = \sqrt{\left ( 1-\frac{m}{r} \right )^2}dt^2[/tex]

    [tex]\frac{dx^2}{dt^2 } + \frac{dy^2}{dt^2 } = \sqrt{\left ( 1-\frac{m}{\sqrt{x^2+y^2}} \right )^2} = c^2[/tex]

    [tex]c= 1-\frac{m}{\sqrt{x^2+y^2}}[/tex]

    [tex]\frac{\partial c}{\partial y}= \frac{m y}{ \left (\sqrt{x^2+y^2} \right )^3 }[/tex]

    which we can integrate with respect to x to find the total deflection.
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