I'm not assuming it, it's the conclusion I draw from your unwillingness to acknowledge what Einstein actually said.
The only thing you can conlude from that is that I don't just blindly follow Einstein or what you tell me to believe about Einstein. In any case, since I have just
proved to you that I am perfectly capable of investigating for myself whether a quantity in general relativity is coordinate-dependent or not, I've proved your conclusion was wrong.
Yes he does, but a mathematical proof that inhomogeneity of $$g_{\mu\nu}$$ is coordinate-dependent doesn't get to the heart of the matter.
What are you talking about? I've just proved, right in front of you, that the $$g_{\mu\nu}$$ can have zero gradient in one coordinate system and non-zero gradient in another. That
contradicts the idea that the $$g_{\mu\nu}$$ gradients are phyical, coordinate-independent quantities. I prove that and all you can say is "this isn't an issue"?
It isn't an issue because in extremis one light ray closely parallel to another doesn't "see" that the other light ray bends.
Then you agree with me. Light bending is coordinate-dependent. You've just described a situation in which we can eliminate the bending in the description of light: compare it to another nearby light ray.
But you still can't see the underlying reality.
I wasn't attempting to say anything about the underlying reality. I've just told you that inhomogeneity of the metric components isn't a good candidate for that.
No. Electrons are "made of light" as demonstrated by pair production.
No it doesn't. This is another converse error fallacy. Pair production does not
require electrons to be "made of light". Proof: the standard model is perfectly able to predict pair production, and it doesn't need to assume electrons are made of light in order to accomplish this.
Yes, and that's an equation of motion. Through space. It isn't an equation for curved spacetime. Try showing me where Einstein says it is. When you can't, you might at last appreciate the distinction between Einstein's interpretation of GR and the modern interpretation.
Nope, it's an equation of motion in
space-time. This is
blindingly obvious if you actually follow the derivation of the equation (which, be honest, you haven't, have you?). Throughout Einstein's paper, in all the quantities which have a Greek letter index on them (such as $$x^{\mu}$$) the index runs over four dimensions, including time. The few exceptions are explicitly stated.
In fact, not only is this an equation in space-time, but this is
absolutely necessary in order for GR to be able to reproduce Newtonian gravity. The equation of motion of a particle in a Newtonian gravitational field is:
$$
\ddot{x}^{i} \,-\, g^{i} \,=\, 0\,,
$$
where each dot above a quantity represents a time derivative (so $$\ddot{x}^{i}$$ is an acceleration), and $$g^{i}$$ is a 3-vector representing the acceleration due to gravity (for example its norm is about 9.8 m/s[sup]2[/sup] near the Earth's surface). In Newtonian gravity the acceleration of a particle depends only on the gravitational field $$g^{i}$$. On the other hand the geodesic equation in space is:
$$
\ddot{x}^{i} \,+\, \Gamma^{i}_{jk} \dot{x}^{j} \dot{x}^{k} \,=\, 0\,.
$$
Here the acceleration of a particle depends not only on the values of the $$\Gamma^{i}_{jk}$$, but also on its velocity $$\dot{x}^{i}$$. If matter obeyed this equation, a particle at rest would not accelerate due to a gravitational field, and it's acceleration would increase quadratically with velocity. This is not what Newtonian gravity predicts, so the geodesic equation just in space is incapable of reproducing Newtonian gravity. On the other hand, the geodesic equation in space-time is
$$
\ddot{x}^{\rho} \,+\, \Gamma^{\rho}_{\mu\nu} \dot{x}^{\mu} \dot{x}^{\nu} \,=\, 0 \,,
$$
where, this time, all the indices take on four values including time, and a dot (like in $$\dot{x}^{\rho}$$) indicates a
proper time derivative. For non-relativistic velocities, $$\dot{t} \equiv \frac{\mathrm{d}t}{\mathrm{d}\tau} = \gamma \approx 1$$, and $$\dot{x}^{0} \equiv c \dot{t} \approx c$$. So if you can arrange things in such a way that the Christoffel symbols $$\Gamma^{i}_{00}$$ are the only non-zero ones, then with this approximation the geodesic equation for the space-like components reduces to
$$
\ddot{x}^{i} \,+\, \Gamma^{i}_{00} c^{2} \,=\, 0 \,,
$$
and you recover the Newtonian equation of motion just with the identification $$\Gamma^{i}_{00} c^{2} = - g^{i}$$.
This is how you recover Newtonian gravity from the geodesic equation. You
specifically need the time-like components in the geodesic equation for this to work. Einstein knew this. He gives his own derivation of this equation in section 21 entitled "Newton's Theory as a First Approximation" of
his paper. The equation I posted above appears in a few different forms on pages 194 and 195. Notice how he keeps referring to components like $$\Gamma^{\tau}_{44}$$, $$[44,\,\tau]$$, and $$g_{44}$$. He does this because he's working in a four dimensional space. Throughout Einstein's paper, 4 designates the time-like component (it's more conventional to label it 0 nowadays). This is explicitly stated on page 151, and he explicitly states that he's working in space-time at the end of page 153 as well as frequently alluding to it in section 4. This is important since it's in these sections that Einstein establishes the conventions he uses throughout the rest of his paper.
Also, in case you're wondering, the reason I have a factor of
c[sup]2[/sup] and Einstein doesn't is because Einstein followed the convention of working in units where
c = 1, as stated in a footnote on page 154.
So you're completely wrong when you say the geodesic equation is just an equation in space. As I've just shown, if it were, GR wouldn't even be able to recover Newtonian gravity.
Not at all. But I am still wondering why you can't see that it's the motion that's curved.
I've already explained that to you:
the concept is coordinate-dependent.
Yep. Because I've shown you that we're made of the damn stuff.
No you haven't. You've used the same sort of argument you always do: you give the example of pair production and then
choose to interpret it a certain way. That doesn't prove anything. And this is pretty irrelevant to the speed of light being coordinate dependent.
But to answer the question: you get the formula for the speed of light that you copied earlier from the weak field approximation, where you set
$$
\mathrm{d}s^{2} \,=\, -\, c^{2}(1 + h) \mathrm{d}t^{2} \,+\, \mathrm{d}x^{2} \,+\, \mathrm{d}y^{2} \,+\, \mathrm{d}z^{2} \,,
$$
where $$h = 2\Phi/c^{2}$$ and $$\Phi$$ is the Newtonian potential (this is the metric you use to recover Newtonian gravity, by the way). Set $$\mathrm{d}s = 0$$ and you find that the "speed" of light is
$$
\sqrt{\Bigl(\frac{\mathrm{d}x}{\mathrm{d}t}\Bigr)^{2} \,+\, \Bigl(\frac{\mathrm{d}y}{\mathrm{d}t}\Bigr)^{2} \,+\, \Bigl(\frac{\mathrm{d}z}{\mathrm{d}t}\Bigr)^{2}} \,=\, c \sqrt{1 \,+\, h} \,=\, c \sqrt{1 \,+\, 2\Phi/c^{2}} \,\approx\, c ( 1 \,+\, \Phi/c^{2} ) \,,
$$
which is the equation you copied in one of your posts above. (See? Not only do I know this result, I also know where it comes from.)
Now pick a certain point $$\bar{x}_{0}$$ in space, and let $$h_{0} = 2 \Phi_{0} / c$$ be the value of $$h$$ at that point. Then do the coordinate substitution $$t \mapsto \sqrt{1 + h_{0}} t$$, and the speed of light in this new coordinate, at the location $$\bar{x}_{0}$$, is
c again.
Another way of seeing the same result is that the Newtonian potential is only defined up to an overall constant, so you can pick any point in space as your origin and choose to set $$\Phi = 0$$ there by convention, which according to the above formula makes the speed of light
c there. This is equivalent to what I did above since, as a first approximation, the rescaling $$t \rightarrow \sqrt{1 + h_{0}} t$$ is mathematically equivalent to resetting $$\Phi \rightarrow \Phi - \Phi_{0}$$.
The fact that the speed of light is always
c in a locally inertial coordinate system, by the way, is trivially true by the fact that a locally inertial coordinate system is
defined as one in which
$$
\mathrm{d}s^{2} \,=\, -\, c^{2} \mathrm{d}t^{2} \,+\, \mathrm{d}x^{2} \,+\, \mathrm{d}y^{2} \,+\, \mathrm{d}z^{2} \;,
$$
locally, and in which all metric derivatives are zero at a point.
You're the one doing this, not me.
What are you talking about?
You are presenting an explanation for something and then committing the fallacy of acting like that rules out any other possible explanation. This is a textbook example of completely broken logic. $$P \rightarrow Q$$ does not imply $$Q \rightarrow P$$. It's you, not me, who is resorting to such arguments. Practically every time you claim some evidence "proves" some explanation you give, you are committing this fallacy.
Aaagh! WMAP tells us space is flat. So that family of solutions is wrong.
No, they're
not being realized. I can't believe you still don't understand the difference. And this doesn't in any way address the point I raised: Friedmann, Lemaitre, Robertson, and Walker started with the assumption of
homogeneity of
space and ended up with the
sphere, without flinching. How do you explain that if Friedmann, Lemaitre, Robertson, and Walker didn't consider the sphere homogenous? I just want to understand how you can believe you agree with these guys when they obviously considered the sphere homogenous and you don't. You don't seem able to give a straight answer to this.
Yes I did, and it's clear that the crucial point didn't sink in. We define the second using the motion of light. And the metre. That's why we always measure the local speed of light to be the same old 299792458 m/s.
No, it doesn't explain anything. Michelson and Morley did not use those definitions of the metre and the second. At the time the standard metre was defined as the length of a bar of platinum kept in France. They still got an invariant
c. So the invariance of
c doesn't depend on us defining our units in any particular way.
The ruler is made of matter. That's made of electrons and protons. And we can make electrons out of light in a lab. And we can turn protons and antiprotons into light in a lab. So your ruler is made of light too.
So point to a mathematical model that 1) reproduces all the predictions of the standard model and 2) models fermions as made of light. Without such a model, don't make such grand claims.
By the way you're making a similar fallacy to the one Robert Close makes, which I explained [POST=2698800]here[/POST]. Measuring invariance of
c does not require everything to be made of waves obeying the massless wave equation. It is sufficient if matter is governed by physics whose mathematical expression is Lorentz invariant. The standard model is a case in point of this. The whole thing, and not just the electrodynamic sector, is Lorentz invariant. Since we're made of stuff governed by the standard model, this alone predicts that we will measure an invariant
c just using rods and clocks, no matter the details of their internal composition.
I read it. And I understand it perfectly. The "explanation" is garbage.
Well it looks fine to me. So I can only conclude you
didn't understand it. Do you know what a null vector is?
And this person, be it Carlip or Gibbs, ends up saying Finally, we come to the conclusion that the speed of light is not only observed to be constant; in the light of well tested theories of physics, it does not even make any sense to say that it varies. It varies, pryzk. Einstein was right, and those light clocks prove it. Clocks clock up motion, not "the passage of time". That's what clocks do. And when one light clock goes slower than the other, it's because the light goes slower. Just replace the clocks with tape reels to get it. Two identical tapes rolling in tamper-proof boxes, sychronised by you and handed to me. I bring them back a while later and one's wound 1000m of tape and the other one 999m. Not because the flow of time was reduced. Because one tape was rolling slower. That's the end of it.
I've just shown above that I can locally make the speed of light anything I want it to be just by rescaling the time coordinate. In fact I don't know why you pointed out that
c is invariant by definition according to the modern definiton, because that's also detrimental to your position: if we can make the speed of light invariant by definition, we're obviously not forced to accept that it varies.
No, I'm not. Instead you're still ducking this issue. Now come on, address it. I want your next post to address either those light clocks or those tape reels, and nothing else. Anything else is an evasion. Come back to the other points after you've demonstrated that you are not ducking this issue.
How many times do I need to explain that, because you yourself said "evidence doesn't distinguish between interpretations",
I have nothing to duck. I'm not going to address
anything until
you present a coherent case. I originally confronted you with the fact that all the recent evidence for GR tested the modern interpretation. You wormed your way out of that with "evidence doesn't distinguish between interpretations". So either explicitly take that back and
properly address the point I made, or quit complaining when I throw your own excuse back at you.
All I'm asking is for you to make your mind up what your position is and stick with it. I'm not going to subscribe to a double standard. Can evidence distinguish between interpretations of GR or not? Pick one and then try to act like you actually believe what you said.
Anything else is evasion. Come back when you show you're
not ducking the fact you're trying to set a double standard.