# Photon Propagation : Straightline or Helix ?

Finding it difficult to commit ?
No, Unless you have a problem in english?
Is it straightline or non straightline ?? You have no third choice, here...
Let me explain: You fire that bullet was it? parallel to the ground, OK, from the moment it leaves the barrel, gravity is acting on it>
It want's to go in s straight line, but due to gravity is deviated from that path.
Which again supports the first factual answer you were given by origin at post 3 or 4..everything travels in a straight line unless acted on by forces, or if you prefer GR, unless affected by curved spacetime......which from memory you have denied exists...am I correct?

Good that you attempted some content contribution, but unfortunately it is not correct in the context. Rpenner has softly nudged you but I will be blunt here...
Whatever nudge rpenner has given anyone, it pales into insignificance with the bump he has given you.
Let me again sum of the gist of this thread:
Everything will travel in a straight line unless acted on by contrary forces [Newtonian] or is influenced by the geometry of spacetime. [GR]
When you finally accept that oh so obviously correct answer, then you may salvage whatever reputation you have left.
Goodnight, I need my beauty sleep.

No, Unless you have a problem in english?

Let me explain: You fire that bullet was it? parallel to the ground, OK, from the moment it leaves the barrel, gravity is acting on it>
It want's to go in s straight line, but due to gravity is deviated from that path.
Which again supports the first factual answer you were given by origin at post 3 or 4..everything travels in a straight line unless acted on by forces, or if you prefer GR, unless affected by curved spacetime......which from memory you have denied exists...am I correct?

Straightline or Non- straightline ?

You are obfuscating and typing nonsense..

Please don't post nonsense on the science subforums.
Let me explain: You fire that bullet was it? parallel to the ground, OK, from the moment it leaves the barrel, gravity is acting on it>
It want's to go in s straight line, but due to gravity is deviated from that path.
Which again supports the first factual answer you were given by origin at post 3 or 4..everything travels in a straight line unless acted on by forces, or if you prefer GR, unless affected by curved spacetime......which from memory you have denied exists...am I correct?

But where is the force on bullet ?

Things regarding projectile motion is known to almost all the science graduates, no big deal.
But your above post is in total mess...first you say that it will travel in geodesic, you piggyback others that geodesic is straightline, then you say it will try to go in straightline, then you say it will be deviated through gravity..somewhere you say gravity is force...somewhere else you say no force then straightline....implying that its path is non-straightline.........you have assiduously avoided answering in single word...Straightline or no-straightline..Why ?

This confirms that your stand is just parrotized nonsense, without actually understanding the significance of topic in hand.

Good that you attempted some content contribution, but unfortunately it is not correct in the context. Rpenner has softly nudged you but I will be blunt here...
umm, i would suggest comprehending rpenner's post. (shakes head)

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Whatever nudge rpenner has given anyone, it pales into insignificance with the bump he has given you.
Let me again sum of the gist of this thread:
Everything will travel in a straight line unless acted on by contrary forces [Newtonian] or is influenced by the geometry of spacetime. [GR]
When you finally accept that oh so obviously correct answer, then you may salvage whatever reputation you have left.
Goodnight, I need my beauty sleep.
Paddoboy -- perhaps it is your lack of sleep but you have been terrible at summarizing GR and terrible at supporting your points in persuasive manner. It doesn't matter if you are conveying your best understanding of this subject when you don't have more than a gloss of it and sometimes are wrong. It's not my job to correct every mistaken claim, but when people shroud themselves in ego and assumed positions of righteousness, it makes my job harder. The God may be more interested in tweaking your nose than arguing like a seeker of knowledge. So you need to be a responsible adult and support your points better without having to post multiple times per hour.

The issue has always been one of definitions. You can't argue with definitions. You can't switch back and forth between different definitions of straight line and force in the same discussion because you won't be talking about the same thing. That's equivocation, which should be called out as a foul.

Everything will travel in straight lines unless acted on by a net non-zero force. That's very nearly the definition of force in terms of straight lines. So what is the definition of straight line? The God hasn't come close to engaging this basic question, so probably is a troll. Farsight won't even engage with the subject even when presented with Einstein's formal definitions, which is pretty damn ridiculous given his modus operandi of running to Einstein's papers as the purported sole source of authority.

• Newton assumed absolute Euclidean space and absolute Euclidean time so his straight lines are those of Cartesian coordinates where $$x''(t) = 0, y''(t) =0, z''(t) = 0$$. Newton was able to unify terrestrial and celestial motions by assuming gravity was a force. Later people would describe electromagnetism as a force and wonder why inertial mass and gravitational charge were one and the same.

• Einstein, in Special Relativity, rejected separate space and time and connected them into a structure of a new sort of geometry where the straight lines in Cartesian coordinates are $$x''(\lambda) = 0, y''(\lambda) =0, z''(\lambda) = 0, t''(\lambda) =0$$ but the ones available to matter those where : $$(ct')^2 - (x')^2 - (y')^2 - (z')^2 > 0$$. These two combine to give basic agreement on what are straight lines with Newton, but now the coordinate velocity of material particles are limited to be below the speed of light in vacuum. We get the twin paradox in Special Relativity because a straight line is the longest path available to material particles, for the metric of this geometry is basically the same thing as proper time. But Special Relativity could not be tweaked to give a satisfactory theory of gravity.

• Einstein, in General Relativity, went to 19th century descriptions of manifolds and tweaked them to have the same local geometry as special relativity. Now Cartesian coordinates are much less special because it is geometry that dictates what a straight line is. The tool that connects coordinates and geometry is the symmetric metric, $$g_{\mu\nu}$$, but unlike earlier manifolds, now it is not positive-definite. The definition of a straight line is now
$$\forall \mu \in \{ 0,1,2,3 \} \; \sum \limits_{\nu \in \{ 0,1,2,3 \} } g_{\mu\nu} x''^{\nu}(\lambda) + \sum \limits_{\alpha, \beta \in \{ 0,1,2,3 \} } \frac{1}{2} \frac{ \partial g_{\beta\mu} }{ \partial x^{\alpha} } x'^{\alpha}(\lambda) x'^{\beta}(\lambda) + \frac{1}{2} \frac{ \partial g_{\mu\alpha} }{ \partial x^{\beta} } x'^{\alpha}(\lambda) x'^{\beta}(\lambda) - \frac{1}{2} \frac{ \partial g_{\alpha\beta} }{ \partial x^{\mu} } x'^{\alpha}(\lambda) x'^{\beta}(\lambda) = 0$$​
which is painful in coordinate-based analysis, but in geometric language can be as simple as $$\mathbf{\nabla _ u u} = 0$$ which says a straight line is compatible with parallel transport on a curved manifold of its own tangent vector. Likewise the straight lines available to material particles are those where $$\sum \limits_{\mu, \nu \in \left\{ 0,1,2,3 \right\} } g_{\mu\nu} x'^{\mu} (\lambda) x'^{\nu} (\lambda)$$ has the correct sign (which is a matter of convention for defining the metric). Now, in the math and philosophy of general relativity, the phenomena of gravity is not described as a force, but as how the content of the universe shapes its geometry.

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I disagree, Geodesic is not the shortest between two points in curved spacetime, null geodesic is.
That's nonsense. Not all pairs of events can be connected by null geodesics because that means that light could propagate from one to the other. So that's not helpful in describing paths taken by material particles.

While often the variational principle finds the smallest value of a quantity, it's actually targeting the local extrema. So it may be a local minimum instead of the global minimum or it may even be a maximum. In SR, the path with the longest proper time between time-like separated events is the straight line. I think I have run out of time to demonstrate that.

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Paddoboy -- perhaps it is your lack of sleep but you have been terrible at summarizing GR and terrible at supporting your points in persuasive manner. It doesn't matter if you are conveying your best understanding of this subject when you don't have more than a gloss of it and sometimes are wrong.
The principals and ideas that I'm trying to convey are generally correct: If they weren't you would say so and others would illustrate my error with reputable links.
Sorry that I'm not the expert mathematician that you are.
The God may be more interested in tweaking your nose than arguing like a seeker of knowledge.
Nice to see that recognised. And is it not a form of trolling?
The God hasn't come close to engaging this basic question, so probably is a troll. Farsight won't even engage with the subject even when presented with Einstein's formal definitions, which is pretty damn ridiculous given his modus operandi of running to Einstein's papers as the purported sole source of authority.
Yep, that's the bones of contention!

.
Sorry that I'm not the expert mathematician that you are.
you can be-- remember that.

you can be-- remember that.

Thanks krash, too old at this stage of life I think.

It's not my job to correct every mistaken claim, but when people shroud themselves in ego and assumed positions of righteousness, it makes my job harder.

I do not shroud myself in ego, I'm doing my best to deflate the agenda driven ego of others that claim they can rewrite 21st century cosmology on a science forum.
Perhaps my knowledge is not much above what some of our trolls deride as "pop science" but the inane stupidity in believing they are the apparent Saviour of 21st century cosmology is what I'm trying to show in my own small way.

I also recognise the fact that the nature of this forum in allowing far more leniency for our many trolls, means that when it is time to bring them into gear, the message that "justice must not only be done, it must be seen to be done" and the usual accusations of bias from these same trolls against the mods, may be the impetus on your rather derisive post to me.
That's OK, I do not have any inflated ego to bruise.
And I certainly do not assume any position of righteousness and would have thought that to be the position of my adversary along with of course his arrogance.
Which is why on most occasions I will always give a reputable link showing its not my position, but the generally accepted position of mainstream cosmology.

I also as I have said many times recognise the difficulty that the mods/admins have in doing their job, along obviously with the sacrifice of their time and responsibility, and I'm sorry that your job is made harder.
Again my point is the message and principal involved in my refutations of what I see as purely and simply wrong.
You mentioned re my adversary in this "debate" how he is obfuscating and moving the goal posts whenever he is cornered....Yet that continues.

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Again, geodesics are straight lines and we were talking about all bodies in curved spacetime including light..
Again, this is popscience nonsense. Curved spacetime relates to the tidal force. There is no detectable tidal force in the room you're in. Little g near the floor is exactly the same as g up by the ceiling. But your pencil still falls down. In similar vein light still curves even when there's no detectable spacetime curvature. Just like the path of the marble curves on the non-curved tilted board.

Fourth warning. Please don't misteach physics on the science subforums.
If you mentally replace the rigid tilted board with a large sheet of fairly thin plastic, you can envisage lifting one end of it. The sheet is now curved. If this curvature wasn't present, the whole sheet would be flat and horizontal. But when you roll your marble, it doesn't "follow the curvature of the sheet". Its path curves because the sheet is tilted.

In similar vein when light curves in a gravitational field, it doesn't "follow the curvature of spacetime". Instead it curves because of the spacetime tilt, which is in turn because of the spacetime curvature, which is in turn because a concentration of energy in the guise of a massive star conditions the surrounding space, this effect diminishing with distance as per the rubber-sheet depictions.

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Curved spacetime relates to the tidal force. .
This, Farsight, is 100% wrong.

The so-called tidal force relates to the gravitational gradient, a first-order differential operator that maps vector fields onto scalar fields.So what is the GR analogue of the gradient as it appears in Newton gravity? Tell us please

On the other hand, curvature is the GR analogue of the divergence of the gradient. This is commonly referred to as the Laplacian, and is a second-order differential operator that maps vector fields onto vector fields (or scalar fields onto scalar fields). So what is the GR analogue of the Laplacian?

Failure to answer these simple questions will seriously damage your credibility on this forum

Curved spacetime relates to the tidal force.
Not quite. You are mangling similar-sounding terms. Spacetime curvature relates to the tidal force as we are talking about differential accelerations of nearby nearly parallel geodesics.

That was demonstrated in Post #78 where the differential acceleration was calculated as proportional to the Riemann curvature tensor.

Naturally, if we have spacetime curvature being non-zero, then the geometry of space-time is not flat. So we have reason to talk about curved spacetime. In curved spacetime, even if we choose coordinates so that the local geometry has the Minkowski metric, we cannot make the second derivatives of the metric go away. So the definition of straight line differs from those of Minkowski space. A pencil falling? A satellite in orbit? The bending of starlight which grazes the sun? All of these trajectories are examples of geodesics of a nearly Schwarzschild geometry.
http://www.physics.usu.edu/Wheeler/GenRel/Lectures/GRNotesDecSchwarzschildGeodesicsPost.pdf

It's like the difference between chocolate milk and milk chocolate, and confusing them makes you look like a non-native speaker. Spacetime curvature is a local property of spacetime, which may be zero (or negligible). Curved spacetime is a spacetime with that property being not everywhere non-zero. Even slightly curved spacetime can have macroscopic effects over macroscopic scales. Pluto takes nearly 250 years to close its orbit, but it is still on a geodesic trajectory.

Additionally, to describe the pencil as falling is a parochial choice of coordinates not based on natural geodesics of spacetime but on a coordinate system where vertical position is in relation to the floor. So first-order gravitational effects are not based on some "tilt" but instead on conflict between coordinate systems and straight lines. In other words, the geodesic equation in Post #41, Post #176, and Post #206.

Finally, both "tilt" and "rubber sheet" attempt to explain gravity in terms of gravity. That's an analogy going around in circles. It's better just to use the math of general relativity. like that from Post #176.

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The so-called tidal force relates to the gravitational gradient, a first-order differential operator that maps vector fields onto scalar fields.
I don't follow.

If V(r) = - GMm/|r| , then $$F(\vec{r}) = - \nabla V(r) = - \frac{GMm}{r^3} \vec{r} \propto \frac{1}{r^2}$$ while tidal force is $$| \vec{s}| \times \nabla_{\vec{s}} F(\vec{r}) \approx - \frac{GMm}{| \vec{r} + \vec{s} |^3} ( \vec{r} + \vec{s} )+ \frac{GMm}{r^3} \vec{r} \propto \frac{1}{r^3}$$ in Newtonian gravitation.

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Not quite. You are mangling similar-sounding terms. Spacetime curvature relates to the tidal force as we are talking about differential accelerations of nearby nearly parallel geodesics.

Would I be right in phrasing that as...tidal 'forces' are evidence of spacetime curvature. In other words, the spacetime geometry over an area may vary so as to give the appearance of different accelerations of two particles. Newton would say, this is the result of different forces acting in that area, and hence the term tidal 'force'.

Would I be right in phrasing that as...tidal 'forces' are evidence of spacetime curvature. In other words, the spacetime geometry over an area may vary so as to give the appearance of different accelerations of two particles. Newton would say, this is the result of different forces acting in that area, and hence the term tidal 'force'.
There's plenty more evidence of spacetime curvature. Every phenomenon associated with gravitation is in GR explained by spacetime curvature and therefore forms evidence for it. But precision tests are the best evidence for it. Tidal forces near the Earth have never been measured on a scale, but have shown up in separation of orbits of spacecraft, and of course, the tides. Atomic clocks ticking at different rates on different floors is an example of a precision test we can do.

I don't follow.

If $$V(r) = - GMm/|r|$$ , then $$F(\vec{r}) = - \nabla V(r) = - \frac{GMm}{r^3} \vec{r} \propto \frac{1}{r^2}$$ while tidal force is $$| \vec{s}| \times \nabla_{\vec{s}} F(\vec{r}) \approx - \frac{GMm}{| \vec{r} + \vec{s} |^3} ( \vec{r} + \vec{s} )+ \frac{GMm}{r^3} \vec{r} \propto \frac{1}{r^3}$$ in Newtonian gravitation.
OK, if I understand you correctly, we are using the term "tidal force" rather differently.

I was using this term literally - the fact that, say Moon disproportionately exerts a gravitation effect on the proximal face of Earth as compared to the distal face.

If your $$s$$ refers to spacetime separation, you seem to be say something like this......

Given 2 test particles equidistant from the "centre" of a gravitational source, then the "force" they experience will not only be "directed" to the gravitational centre, but also appear to be "directed" toward each other. This is a common pop-sci explanation of spacetime curvature

I have a slight problem with this........

OK, if I understand you correctly, we are using the term "tidal force" rather differently.

I was using this term literally - the fact that, say Moon disproportionately exerts a gravitation effect on the proximal face of Earth as compared to the distal face.
Given by the third expression in Newtonian gravitation $$- \frac{GMm}{| \vec{r} + \vec{s} |^3} ( \vec{r} + \vec{s} )+ \frac{GMm}{r^3} \vec{r} = F(\vec{r} + \vec{s}) - F(\vec{r})$$.

If your $$s$$ refers to spacetime separation, you seem to be say something like this......
Nope, it's just a vector as per Newton.

Given 2 test particles equidistant from the "centre" of a gravitational source, then the "force" they experience will not only be "directed" to the gravitational centre, but also appear to be "directed" toward each other. This is a common pop-sci explanation of spacetime curvature

I have a slight problem with this........
It's saying in the limit of small $$\vec{s}$$ that the tidal force tends to separate trajectories/orbits/material bodies if $$\vec{s} \parallel \vec{r}$$ and tends to squeeze them together if $$\vec{s} \perp \vec{r}$$.

$$\lim \limits_{h \to 0} \frac{ F(\vec{r} + h \vec{s}) - F(\vec{r}) }{ h } = \frac{3 GMm ( \vec{s} \cdot \vec{r} ) }{r^5} \vec{r} - \frac{GMm}{r^3} \vec{s} \\ \vec{s} \cdot \vec{r} = 0 \Rightarrow \lim \limits_{h \to 0} \frac{ F(\vec{r} + h \vec{s}) - F(\vec{r}) }{ h } = - \frac{GMm}{r^3} \vec{s} \\ \vec{s} \propto \vec{r} \Rightarrow \lim \limits_{h \to 0} \frac{ F(\vec{r} + h \vec{s}) - F(\vec{r}) }{ h } = \frac{2 GMm}{r^3} \vec{s}$$​

in Newton's Universal Gravitation.

While in the Schwarzschild solution of General Relativity, the analogous calculation is $$R_{\beta\gamma\delta}^{\alpha} T^{\beta} T^{\gamma} X^{\delta}$$ where only 24 of the 256 components of the curvature tensor are non-zero, and in closest analogy T is time-like, X is space-like. So lets fix T to be in the direction of the time coordinate and $$T^2 = g_{\alpha \beta} T^{\alpha} T^{\beta}= -c^2$$.

In this case, only three non-zero components of the curvature tensor are used. I get
$$T^{\alpha} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \\ 0 \\ 0 \\ 0 \end{pmatrix} \\ R_{t t \delta}^{\alpha} = \begin{pmatrix} 0 & 0 & 0 & 0\\0 & \frac{r_{s}}{r^{4}} c^{2} \left(r - r_{s}\right) & 0 & 0\\0 & 0 & - \frac{ c^{2} r_{s}}{2 r^{4}} \left(r - r_{s}\right) & 0\\0 & 0 & 0 & - \frac{c^{2} r_{s}}{2 r^{4}} \left(r - r_{s}\right)\end{pmatrix} \\ R_{\beta\gamma\delta}^{\alpha} T^{\beta} T^{\gamma} X^{\delta} = \frac{c^2 r_s}{2 r^3} \begin{pmatrix} 0 \\ 2 X^{r} \\ - X^{\theta} \\ - X^{\phi} \end{pmatrix}$$
which is precisely what you would expect if $$t$$ and $$r$$ in Schwarzschild coordinates meant exactly the same thing as $$t$$ and $$r$$ in Newtonian physics.

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