# Has any Prominent Physicist Ever Admitted to Not Understanding Magnetism?

Discussion in 'Physics & Math' started by Eugene Shubert, Sep 17, 2015.

1. ### rpennerFully WiredValued Senior Member

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That's just evidence that Einstein didn't understand all the ramifications of General Relativity as well as mainstream post-graduate physicists do today.

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3. ### Farsight

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He did, but they don't. See this:

5. ### Farsight

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Or how about this by Professor Ned Wright of UCLA: "In a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light. The figure below shows a bundle of rays passing the Sun at various distances. The rays are always perpendicular to the wavefronts which mark the set of points with constant travel time from the star. In order to bend the light toward the star one needs to delay the wavefront near the star."

7. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

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You should have some conversation.

No one will think better of you if all you do is spam. But hey, maybe you think so much of yourself.

8. ### James RJust this guy, you know?Staff Member

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There are two approaches at work in the explanation of the bending of light by a star.

1. Assume spacetime is flat and the speed of light varies with distance from the star; OR
2. Assume spacetime is curved near the star and the speed of light is constant everywhere.

Approach (1) is roughly equivalent to what you see if you're in a reference frame far from the star. Approach (2) is what you see if you move along through the space that the light moves through.

If you want to work out by how much the speed of light apparently varies in approach (1), you'll need to use general relativity, which explains the apparent variation as an effect of spacetime curvature - which is, of course, approach (2).

So, really, whether you use approach (1) or approach (2), you're really just looking at the same theory two different ways.

9. ### Farsight

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That's pretty reasonable, though IMHO there's a little issue with the phraseology. If you say the speed of light varies, you're also saying spacetime is not flat.

On top of that there's another issue about spacetime being tilted as opposed to curved. See the titled light cones here. The rubber-sheet analogy isn't perfect, but see the Wikipedia Riemann curvature tensor article and look at the depiction on the right. The gedanken marble rolls towards the Earth because of the slope or tilt rather than the curvature. But you need the curvature to have a slope. If there's no curvature your whole plot is flat and level.

Think about a board. It can be flat as in not curved, and it can be flat as in horizontal. When it's tilted it's still isn't curved, but now a marble will roll down it. The room you're in is something like that. Your pencil doesn't fall down because the spacetime in the room you're in is curved, but instead because it's tilted. But it's tilted because the spacetime round the Earth is curved.

Last edited: Oct 20, 2015
10. ### rpennerFully WiredValued Senior Member

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Yes, because it is an analogy which can't be used to extract dynamics. Unlike the rubber sheet, there is no analogue of "slope" in general relativity since there is no direction to point to "up." The important property is not some global definition of "level" and local "slope" variations with respect to that global "level" but local variation from flat geometry, i.e. a metric which changes with respects to space and/or time in a way where the Christoffel symbols indicate that parallel transport of vectors is path-dependent, i.e. space-time curvature.

In general curvilinear coordinates where $x^0 \equiv t$ is coordinate time, we have this equation of motion for a test particle:

$\frac{d^2 x^\mu}{dt^2} = \sum_{\alpha,\beta \in \{0,1,2,3\}} \left( \Gamma_{\alpha \beta}^0 \frac{d x^{\mu}}{dt} - \Gamma_{\alpha \beta}^{\mu} \right) \frac{d x^{\alpha}}{dt} \frac{d x^{\beta}}{dt} = \sum_{\alpha,\beta \in \{0,1,2,3\}} \left( \sum_{\nu \in \{0,1,2,3\}} \frac{g^{\nu 0} \frac{d x^{\mu}}{dt} - g^{\nu \mu} }{2} \left( \partial_{\alpha}g_{\nu\beta}+\partial_{\beta}g_{\nu\alpha}-\partial_{\nu}g_{\alpha\beta}\right) \right) \frac{d x^{\alpha}}{dt} \frac{d x^{\beta}}{dt}$

So in the simple case for a massive particle at coordinate rest, where $\left( \frac{d x^{0}}{dt}, \frac{d x^{1}}{dt}, \frac{d x^{2}}{dt}, \frac{d x^{3}}{dt} \right) = (1, 0, 0, 0)$ we have

$\frac{d^2 x^0}{dt^2} = 0 \\ \frac{d^2 x^\mu}{dt^2} = - \frac{1}{2} \left[ g^{0 \mu} \partial_{0}g_{0 0} + g^{1 \mu} \left( 2 \partial_{0}g_{1 0}-\partial_{1}g_{0 0}\right) + g^{2 \mu} \left( 2 \partial_{0}g_{2 0}-\partial_{2}g_{0 0}\right) + g^{3 \mu} \left( 2 \partial_{0}g_{3 0}-\partial_{3}g_{0 0}\right) \right] ; \quad \mu \in \{1,2,3\}$

In Minkowski space, $\partial_{\alpha} g_{\beta \gamma} = 0$ so $\frac{d^2 x^\mu}{dt^2} = 0$.

In a Schwarzschild vacuum, which is the first vacuum solution because it was spherically symmetric and static with respect to the time coordinate, with $(x^0, x^1, x^2, x^3) = ( t, r, \theta, \phi), g_{00} = \frac{r - r_s}{r} c^2, g_{11} = - \frac{r}{r - r_s} , g_{22} = - r^2, g_{33} = - r^2 \sin^2 \theta$ the metric, its derivatives and its inverse are all conveniently diagonal, allowing us to express for a test particle in coordinate-rest as:

$\frac{d^2 r}{dt^2} = - \frac{1}{2} \left[ \frac{r - r_s}{r} \left( \frac{\partial \quad}{\partial r} \frac{r - r_s}{r} c^2 \right) \right] = - \frac{G M}{r^2} \left( 1 - \frac{ r_s }{ r} \right)$

For the Earth, $r_s = \frac{2 GM}{c^2} \approx 0.00887 \, \textrm{m}$, so this is a very small departure from Newton's description of the behavior of gravity. The discrepancy gets larger for bodies at high coordinate speed.

Last edited: Oct 20, 2015
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11. ### OnlyMeValued Senior Member

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The problem with your tilted board.., is that the only way the marble rolls anywhere on a flat board, is when it is slopped or tilted, relative to in the case of gravitation and GR, curved spacetime.

12. ### Farsight

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Yes there is, it's the change in the gravitational potentials. You simply place optical clocks throughout an equatorial slice through the Earth and the surrounding space, then plot the clock rates. You depict lower slower clocks as lower down in a 3D image, and higher faster clock rates higher up. What your plot looks like, is this:

CCASA image by Johnstone, see Wikipedia

That's a picture from the Wikipedia Riemann curvature tensor article. It's a reasonable low-dimensional depiction of curved spacetime. And because it's derived from optical clock rates, it's a plot of the what Einstein and Shapiro referred to as the speed of light. You would call it the coordinate speed of light. The force of gravity of some location depends on the slope of the plot, not how curved it is at that location. And the crucial point to note is that the speed of light does not vary because your plot of the speed of light is curved. It does not vary because "space-time is curved". Instead it varies because a concentration of energy conditions the surrounding space, and this effect diminishes with distance in a non-linear fashion. That metric is not space, it's an abstract thing derived from your measurements, made using things such as optical clocks.

No! And I will reiterate: your pencil does not fall down because of space-time curvature in the room you're in. The latter relates to the tidal force, not the force of gravity. There is no detectable tidal force in the room you're in, and so no detectable space-time curvature. But pick up your pencil, then let go. See how it falls? That's detectable.

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14. ### DaeconKiwi fruitValued Senior Member

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You're two years late.

15. ### Eugene ShubertValued Senior Member

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Even God has delayed His final judgment but I'm not going to deny the justice that will eventually be dispensed at the great, final, awesome settling of destiny.