[Continuation] Why James Webb Telescope rewrites/doesn't the laws of Physics/Redshifts (2)

[W.D.Clinger]

The author of the paper seems to think so. Though I'm not sure.

The opinion of an author who makes mistakes at the level of first-year calculus isn't worth very much.

But it can't surprise too many people here that the author and sole proponent of Helland physics continues to cite and to quote false conclusions that are known to have been derived from specific mistakes in first-year calculus.

So you just threw in a paper you were not advocating, in order to waste the time of your critics?

At this point, anyone who bothers to look at a paper that has been recommended by that cargo cultist knows what they're getting into.

For myself, as a non-physicist who is many decades removed from taking or teaching calculus, the mathematical exercise of performing basic sanity checks on physics papers that are empirically likely to contain obvious errors ("empirically likely" because we have seen time and again that the author and sole proponent of Helland physics has an uncanny knack for citing and quoting papers that contain obvious errors) is a good way to practice some mathematical skills the author and sole proponent of Helland physics has never even attempted to acquire and apparently never will.

 

[IGMisBestM]

Ok. I took it to mean "dusty" means not metal poor.

I suppose it would be useful to define when the "early" universe was and what constitutes "metal poor". A curve from now to the big bang showing what the predicted dust-to-gas ratio is would be the best way to quantify that. Maybe I can find one.

Early is a relative term, earlier galaxies are more metal poor in general. And by metal poor I mean with respect to the local mass-metallicity relationship. Even accounting for their lower mass, high redshift galaxies are much more metal poor than z=0. There is real evolution which is not compatible with static models. There is real galaxy evolution, in line with the natural prediction of increasing metallicity. JWST has shown many differences between early galaxies and the local universe.
I have no idea why you have veered off into dust-to-gas ratios, there is no "big bang prediction" because it depends entirely on the model of galaxy formation and a model of dust formation.

Can we take this to mean that if we describe the metric as time dilating EM waves at their source, that it would be "wrong", in that context, to describe them as being stretched by the expansion of space as they travel?

The amount of time dilation depends on how much the universe has expanded while the light is traveling, it makes no sense to say it happened at the source. How could the source know how far the light would travel, or how the expansion of the universe would evolve in the future. This is not a choice one can make.

 

[Mike Helland]

Early is a relative term, earlier galaxies are more metal poor in general. And by metal poor I mean with respect to the local mass-metallicity relationship. Even accounting for their lower mass, high redshift galaxies are much more metal poor than z=0. There is real evolution which is not compatible with static models. There is real galaxy evolution, in line with the natural prediction of increasing metallicity. JWST has shown many differences between early galaxies and the local universe.
I have no idea why you have veered off into dust-to-gas ratios, there is no "big bang prediction" because it depends entirely on the model of galaxy formation and a model of dust formation.

I kind of poked around for a resource that may clarify what some of those things are. There's metallicity, dust-to-gas, and dust-to-metal.

I'm assuming metal means anything heavier than helium. And dust has to be made of metal.

The metallicity might be what appears in the stars, while the dust-to-gas might be metal outside of stars? That's not clear. That would also make dust-to-metal somewhat confusing, so that's probably not right.

I see that Fe/H or O/H are used in different articles, and that all three (metallicity, DGR, DTM) are based on this. X+Y+Z=1, where X is H, Y is He, and Z is everything else. And all three obviously start at 0 and make their way up to solar levels.

But what actually makes them different isn't very clear.

The amount of time dilation depends on how much the universe has expanded while the light is traveling, it makes no sense to say it happened at the source. How could the source know how far the light would travel, or how the expansion of the universe would evolve in the future. This is not a choice one can make.

Ok.

So here's a thought experiment. Say you have a simple expanding universe, and a supernovae happens at z=1. We're interested in the peak brightness to when the brightness becomes where it used to be. We'll call that T. Maybe two weeks or something.

Let's say when it hits that time, right as the last photon we're interested in is emitted, expansion stops, and the galaxy stops moving away. The photons travel to us through static space.

They should 2T apart, from start to end, when its observed at z=0.

That seems obvious, right.

Are the photons redshifted? Even the last one?

 

[Ziggurat]

So here's a thought experiment. Say you have a simple expanding universe, and a supernovae happens at z=1.

Ignoring peculiar motion, z is not a measure of distance, it is a measure of how much the universe has expanded since emission. If z = 1, then the universe has expanded to twice the size of what it was at emission. This is important because in your hypothetical scenario, you can't really assign a single z value to your supernova.

We're interested in the peak brightness to when the brightness becomes where it used to be. We'll call that T. Maybe two weeks or something.

Let's say when it hits that time, right as the last photon we're interested in is emitted, expansion stops, and the galaxy stops moving away. The photons travel to us through static space.

If the universe has stopped expanding, then how do you get to z = 1? Anything that happens after the universe stops expanding is at z=0, no matter how far away it is, which means the last photon is at z=0. So either your setup doesn't make sense, or the START of the supernova happens at z=1 (the only nonzero value you mentioned) and the end happens at z=0. In other words, between the start and end of the supernova, the universe expands by a factor of 2 and then stops. Note that the end happens farther away than the start even though it's at lower z. This is why it's important to note that z is not actually a measure of distance.

Now obviously an expansion by a factor of 2 and then a sudden halt isn't a realistic scenario, but that's OK, this is just a thought experiment. Under the assumption that this is what you're asking about (because otherwise your setup makes no sense at all), then the first photons from the start of the supernova would be red shifted by a factor of 2, but the last photons would not be red shifted at all.

They should 2T apart, from start to end, when its observed at z=0.

That seems obvious, right.

Wrong. The first photon would be red shifted by a factor of 2, but the last photon wouldn't be red shifted at all. The total time between first and last photon arriving at the observer would be a function of how the expansion happened between the first and the last. If the expansion is constant during that interval, then the red shift will be continuously decreasing for each subsequent photon, and the total time between the first and the last will be a sort of average of the red shifts of each photon, so something like 1.5T. If you front load more of the expansion, the time interval would be closer to 1T, and if you back load more of the expansion, it will be closer to 2T.

 

[Mike Helland]

If the universe has stopped expanding, then how do you get to z = 1?

Based on the time dilation, it's z=1.

Here's an updated image of the scenario. The green line represents the galaxy's motion through spacetime. Traveling at v=c while the supernova is happening, and then v=0 after.

(Not to scale obviously, the yellow lines should be 2 weeks apart but 7 billion years long.)

The time dilation of the SN is determined only by the v=c part.

If the expansion rate were to double, or even reverse, the effect on the distance the photons have to travel will be the same for all photons. That it stops shouldn't affect the time dilation of the supernova. Assuming the observer's relative velocity was v=c when the supernova happened, it's velocity before or after the photons the we're interested in shouldn't matter.

Right?

The first photon would be red shifted by a factor of 2, but the last photon wouldn't be red shifted at all.

The first photon is in the air by itself for 2 weeks while the universe expands.

The second photon is never in expanding space.

if the second photon has a redshift of 0, the first photon should have a redshift representing a distance of 2 light-weeks, not 7 billion years. Right?

 

[Darat]

I've some strong evidence the universe is cyclical and everything will repeat all other again and again and ....

 

[Ziggurat]

Your diagram makes no sense. What coordinates are you using? I don't think you actually know. And if you think you do, you probably screwed up.

 

[Mike Helland]

Your diagram makes no sense. What coordinates are you using? I don't think you actually know. And if you think you do, you probably screwed up.

Heh. Probably.

Seems straightforward enough, though.

If they were co-moving coordinates the galaxy wouldn't be in motion, so it's not those.

If the galaxy's x spatial coordinate is 1 billion light years now, then in the past it would be smaller, 1/(1+z), or e^H₀t.

So if x is the current distance, and x' is the old distance, http://latex.codecogs.com/gif.latex?x^` = xe^{H_0t}.

x is the comoving coordinate. So what's x' called? That's the coordinates shown.

 

[Ziggurat]

If the galaxy's x spatial coordinate is 1 billion light years now, then in the past it would be smaller, 1/(1+z), or e^H₀t.

That's assuming a constant expansion rate. You are postulating an extremely non-constant expansion rate. Your formula relating distance to z fails spectacularly under such a scenario.

 

[Mike Helland]

That's assuming a constant expansion rate. You are postulating an extremely non-constant expansion rate. Your formula relating distance to z fails spectacularly under such a scenario.

That's true. If expansion were to stop, it wouldn't be expanding at a constant rate.

Are you proud of where you landed on that one?

 

[IGMisBestM]

I kind of poked around for a resource that may clarify what some of those things are. There's metallicity, dust-to-gas, and dust-to-metal.

I'm assuming metal means anything heavier than helium. And dust has to be made of metal.

The metallicity might be what appears in the stars, while the dust-to-gas might be metal outside of stars? That's not clear. That would also make dust-to-metal somewhat confusing, so that's probably not right.

I see that Fe/H or O/H are used in different articles, and that all three (metallicity, DGR, DTM) are based on this. X+Y+Z=1, where X is H, Y is He, and Z is everything else. And all three obviously start at 0 and make their way up to solar levels.

The value in these galaxies is the gas phase metallicity, but metallicity can be estimated anywhere. Note that it is the total amount of heavy elements, you do not add these factors together. They are completely different. The ratios do not necessary evolve monotonically, if a galaxy is depleted of gas the ratio can fall.

 

[Ziggurat]

That's true. If expansion were to stop, it wouldn't be expanding at a constant rate.

Are you proud of where you landed on that one?

Are you proud that you missed it?

Your question is poorly framed, and your graph is nonsense because you missed the obvious.

 

[Mike Helland]

Are you proud that you missed it?

Your question is poorly framed, and your graph is nonsense because you missed the obvious.

When expansion stops, the rate of expansion changes to zero. That's pretty obvious.

Also, since proper distance changes, and the speed of light is constant in proper coordinates, the graph should be properly showing proper coordinates.

 

[Ziggurat]

When expansion stops, the rate of expansion changes to zero. That's pretty obvious.

Also, since proper distance changes, and the speed of light is constant in proper coordinates, the graph should be properly showing proper coordinates.

See, here's where you're wrong. The speed of light isn't constant in proper coordinates, it's position-dependent. As a source approaches the cosmological horizon, the speed of light in proper coordinates at that source drops towards zero. The speed of light is also direction dependent, which should be obvious from your graph, since that's the only possible way you could have your source moving away from you at c.

You have no idea what you're doing.

 

[Mike Helland]

See, here's where you're wrong. The speed of light isn't constant in proper coordinates, it's position-dependent. As a source approaches the cosmological horizon, the speed of light in proper coordinates at that source drops towards zero. The speed of light is also direction dependent, which should be obvious from your graph, since that's the only possible way you could have your source moving away from you at c.

You have no idea what you're doing.

The speed of light in comoving coordinates is c(1+z).

The speed of light in proper coordinates isn't c?

If that's true, in what coordinates is the speed of light actually constant in GR?

 

[Ziggurat]

The speed of light in comoving coordinates is c(1+z).

The speed of light in proper coordinates isn't c?

Nope. And again, that should have been obvious even to you. Seriously, look at your graph again: how would you plot the path of a photon emitted from SN start but moving away from the origin? It would obviously and necessarily have a slope more horizontal than 45 degrees.

If that's true, in what coordinates is the speed of light actually constant in GR?

The speed of light is c anywhere that the local coordinates match Minkowski coordinates. Which you can always do locally, since Minkowski space is always the tangent space to a GR manifold.

You won't understand what I just said, but again, you don't understand what you're doing.

 

[Mike Helland]

Nope. And again, that should have been obvious even to you. Seriously, look at your graph again: how would you plot the path of a photon emitted from SN start but moving away from the origin? It would obviously and necessarily have a slope more horizontal than 45 degrees.

I see.

Well, if it was a z=2 galaxy, its speed would be 2c, so it would be moving faster away from the origin than the light that left it.

Are proper coordinates really even graphable in a static 2d image then?

The speed of light is c anywhere that the local coordinates match Minkowski coordinates. Which you can always do locally, since Minkowski space is always the tangent space to a GR manifold.

You won't understand what I just said, but again, you don't understand what you're doing.

If one had been following this thread, that should make sense.

The metric tensor takes input coordinates, and its output are tangent vectors that basically make up the basis vectors of a tangent space whose origin intersects the input coordinate.

That's kind of funny that we'd say the speed of light is constant in general relativity, when it's only constant in the tangent space, the space you kind of imagine to exist outside the manifold in order to simplify your calculations.

 

[Ziggurat]

I see.

Well, if it was a z=2 galaxy, its speed would be 2c,

Uh... no. You're pulling that out of your ass. Even in the case of uniform expansion (which, again, your thought experiment is explicitly NOT), that's not the relationship between z and recessional velocity.

Are proper coordinates really even graphable in a static 2d image then?

Yes, but not easily, and they don't have 45 degree light cones.

That's kind of funny that we'd say the speed of light is constant in general relativity, when it's only constant in the tangent space, the space you kind of imagine to exist outside the manifold in order to simplify your calculations.

Not really. Coordinates are arbitrary in General Relativity, and so speeds measured in arbitrary coordinates are arbitrary. That's not profound, and it says nothing of significance about the speed of light. But you can always do local measurements no matter where you are, and the local measurements on the speed of light will always produce the same results. That is meaningful.

 

[Ziggurat]

Uh... no. You're pulling that out of your ass. Even in the case of uniform expansion (which, again, your thought experiment is explicitly NOT), that's not the relationship between z and recessional velocity.

I should amend this. You probably didn't pull it out of your ass, but it's still wrong. z = v/c is an approximation that's only valid for small z. It's quite wrong by the time you get to z = 1 and above.

 

[Mike Helland]

Uh... no. You're pulling that out of your ass. Even in the case of uniform expansion (which, again, your thought experiment is explicitly NOT), that's not the relationship between z and recessional velocity.

So there's this:

https://cosmocalc.icrar.org/

Which says the cosmological recession velocity is c=1 at z=3, using pure dark energy parameters. I don't know if I have w[0] and w' set right, both to 0 otherwise I got an error.

I thought it was v = cz = H₀D.

I know that's a first order approximation for models where the Hubble parameter changes, but when it's static, D = zc/H₀, for comoving distane, is the actual result.

Yes, but not easily, and they don't have 45 degree light cones.

Interesting.

Not really. Coordinates are arbitrary in General Relativity, and so speeds measured in arbitrary coordinates are arbitrary. That's not profound, and it says nothing of significance about the speed of light. But you can always do local measurements no matter where you are, and the local measurements on the speed of light will always produce the same results. That is meaningful.

At z=2, the galaxy would be moving at cz = 2c, and the light would be moving at c(1+z) = 3c, so to the galaxy, light is moving at c.

If you subtract the Hubble flow from the galaxy and the light, then v_galaxy = 0, and v_light=c.

If the recessional velocity is not cz, it seems that speed of light - speed of galaxy (ie, c(1+z) - cz) does not equal c.

 

[Ziggurat]

I thought it was v = cz = H₀D.

Only for small z. Plug a small number into that calculator you linked and see.

 

[Ziggurat]

OK, so here's another way to understand why your graph is wrong. We're not doing constant expansion, so we don't need to stick to any specific relationship between velocity and z, and we can just ignore that whole issue.

But if z=1 for the start of our supernova, then wavelength expanded by a factor of 2, and space expanded by a factor of 2. That part is model independent.

And your graph doesn't show space expanding by a factor of 2 from the start of the supernova to any other time shown on that graph.

Now, you COULD make an artificial scenario in which a supernova starts, space expands by a factor of 2, the supernova ends, and space stops expanding. But the space-time diagram of that scenario won't look like your graph. It can't look like your graph.

 

[Mike Helland]

Only for small z. Plug a small number into that calculator you linked and see.

That gives pretty different results than this:

http://www.bo.astro.it/~cappi/cosmotools

(eta I think it has to do with the other one's equation of state. Looks like I crashed it again.)

Using the above one, comoving distance is linear, with Ω_Λ=1 and Ω_M=0.

The equation for comoving distance should simplify to d=zc/H₀ for those parameters.

When you add matter things change, and that equation is only an approximation for small z.

 

[Ziggurat]

That gives pretty different results than this:

http://www.bo.astro.it/~cappi/cosmotools

How can that give a different answer for velocity when it doesn't give velocity at all?

 

[Mike Helland]

How can that give a different answer for velocity when it doesn't give velocity at all?

It gives different answers for comoving distance.

It should be:

http://latex.codecogs.com/gif.latex?d(z) = \frac{c}{H_0} \int_0^z \frac{dx}{\sqrt{\Omega_R(1+x)^4 + \Omega_M(1+x)^3 + \Omega_K(1+x)^2 + \Omega_\Lambda}​

https://www.quora.com/How-do-we-cal...smological-body-thats-really-far-away-from-us

If the parameters are 0, 0, 0 and 1, then you get:

http://latex.codecogs.com/gif.latex?d(z) = \frac{c}{H_0} \int_0^z 1 ~ dz^` = z \frac{c}{H_0}​

 

[W.D.Clinger]

The author and sole proponent confirms once again that he doesn't have the slightest idea of what he's talking about.

The metric tensor takes input coordinates, and its output are tangent vectors that basically make up the basis vectors of a tangent space whose origin intersects the input coordinate.

No. Just no.

That sentence contains four different fundamental errors.

That's kind of funny that we'd say the speed of light is constant in general relativity, when it's only constant in the tangent space, the space you kind of imagine to exist outside the manifold in order to simplify your calculations.

That sentence would not have been written by anyone who actually understands calculus.

 

[Ziggurat]

It gives different answers for comoving distance.

Don't care. We weren't talking about distance, and your toy model for your thought experiment isn't described by those equations anyways. I have zero interest in tracking down where you made a mistake this time.

 

[Mike Helland]

That sentence contains four different fundamental errors.

Whether those errors are relevant, or dependent on one textbook/course over another remains to be seen. With your track record, I can only assume there's a 50% chance the errors amount to nothing more than your personal preferences and mental gymnastics.

So what are they?

 

[Mike Helland]

Don't care. We weren't talking about distance, and your toy model for your thought experiment isn't described by those equations anyways. I have zero interest in tracking down where you made a mistake this time.

Well, if not that one, what equation would you use to calculate recessional velocity given a z, for a set of cosmological parameters?

 

[Ziggurat]

Well, if not that one, what equation would you use to calculate recessional velocity given a z, for a set of cosmological parameters?

Why is that even relevant to your toy model which is completely disconnected from standard cosmological models, and isn't described by any of those parameters?

You're completely incoherent.

 

[Mike Helland]

Why is that even relevant to your toy model which is completely disconnected from standard cosmological models, and isn't described by any of those parameters?

You're completely incoherent.

You said the galaxy's recessional velocity at z=1 isn't v=c.

So what is it, and how did you calculate it?

The scale factor for the thought experiment is:

http://latex.codecogs.com/gif.latex? a(t) = \begin{cases} e^{H_0t} & t < \text{SN end} \\ 1 & t \geq \text{SN end} \end{cases}​

That should have been obvious. (Keep in mind t < 0 for past events.)

 

[Ziggurat]

You said the galaxy's recessional velocity at z=1 isn't v=c.

For the standard model, it isn't. But why do you care? You can make a toy model where you can have any recessional velocity you want at any z you want.

The scale factor for the thought experiment is:

http://latex.codecogs.com/gif.latex? a(t) = \begin{cases} e^{H_0t} & t < \text{SN end} \\ 1 & t \geq \text{SN end} \end{cases}​

That should have been obvious. (Keep in mind t < 0 for past events.)

It wasn't obvious, because your diagram was nonsense and your prior description doesn't match this. And under this scenario, none of the standard cosmology equations for how z relates to anything else are going to be correct anyways.

Furthermore, it's all completely unnecessary anyways. If you've got your time dependent scale factor, then just calculate the scale factor at the start of the supernova, calculate the scale factor at the end of the supernova, and that will tell you everything you need to know quite directly since the scale factor in this toy model doesn't change after that. You don't need z because it's not an input, it's an output. You just need H₀ and the time duration of the supernova. If you pick reasonable values for these, you'll find that z isn't close to 1 (assuming zero peculiar velocity).

 

[W.D.Clinger]

The author and sole proponent confirms once again that he doesn't have the slightest idea of what he's talking about.

The metric tensor takes input coordinates, and its output are tangent vectors that basically make up the basis vectors of a tangent space whose origin intersects the input coordinate.

No. Just no.

That sentence contains four different fundamental errors.

Whether those errors are relevant, or dependent on one textbook/course over another remains to be seen. With your track record, I can only assume there's a 50% chance the errors amount to nothing more than your personal preferences and mental gymnastics.

So what are they?

You knew he'd have to ask, even though his errors have been explained again and again.

1. "The metric tensor takes input coordinates"
   
   No. The inputs to the metric tensor are two vectors. The components of those vectors are not coordinates. (Long-time readers of this thread may recall the hilarity of Mike Helland's multiple wrong guesses as he struggled to define a metric tensor, until finally I asked whether he had considered the possibility that none of his attempts actually defined any kind of tensor at all. One of the reasons his attempts didn't define any kind of tensor at all is that he was writing his alleged metric forms as though the metric tensor's inputs were coordinates rather than vector components. It seems he still doesn't know any better.)
   
   
2. "and its output are tangent vectors"
   
   No. The output of the metric tensor is a scalar. Although it is possible to regard a scalar as the simplest kind of vector, that scalar is definitely not a tangent vector.
   
   
3. "that basically make up the basis vectors of a tangent space"
   
   No. The scalar output of a metric tensor is definitely not a basis vector of some tangent space.
   
   
4. "whose origin intersects the input coordinate"
   
   No. The inputs are not coordinates, the output is not a tangent vector, the output is not a basis vector of a tangent space, and (regardless of the tangent space he imagines the scalar output of the metric tensor to inhabit) it is silly nonsense to speak of the origin of a vector space intersecting a scalar (such as a coordinate).

tl/dr: Although the author and sole proponent of Helland physics often babbles about metric tensors, basis vectors, tangent vectors, and coordinates, he doesn't understand any of those things. As we saw just three days ago, he struggles with first-semester calculus.

 

[Mike Helland]

Let's take the first one.

[*]"The metric tensor takes input coordinates"

No. The inputs to the metric tensor are two vectors.

Say the metric tensor is:

http://latex.codecogs.com/gif.latex?g_{\mu\nu} = \begin{bmatrix} -c^2 & 0 & 0 &0 \\ 0 & a(t)^2 & 0 & 0 \\ 0 & 0 & a(t)^2 & 0 \\ 0 & 0 & 0 & a(t)^2 \end{bmatrix}​

If given the (t, x, y, z) coordinates (0, 0, 0, 0) this evaluates to:

http://latex.codecogs.com/gif.latex?g_{\mu\nu} = \begin{bmatrix} -c^2 & 0 & 0 &0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}​

Given (-10 Gy, 0, 0, 0) you get (approx):

http://latex.codecogs.com/gif.latex?g_{\mu\nu} = \begin{bmatrix} -c^2 & 0 & 0 &0 \\ 0 & 0.5^2 & 0 & 0 \\ 0 & 0 & 0.5^2 & 0 \\ 0 & 0 & 0 & 0.5^2 \end{bmatrix}​

Which, indeed gives you a scalar, which you can apply to the basis for that dimension.

So, it seems to me, the metric tensor takes coordinates as an input, and gives you a field of tensors (psuedo-tensors, I suppose), one for every coordinate, which don't change the direction of the basis vectors, but effectively change their length to a bespoke value for each coordinate.

It seems depending on who wrote the textbook/course, the wording on that may have several variations. But that seems to be the gist.

I await your corrections.

 

[W.D.Clinger]

Which, indeed gives you a scalar, which you can apply to the basis for that dimension.

No one who knows what they're talking about would write "the basis for that dimension."

So, it seems to me, the metric tensor takes coordinates as an input,

No.

and gives you a field of tensors

A metric tensor form can refer to coordinates, because that tensor form is describing a tensor field, not a single tensor. By plugging in the coordinates of a point, you obtain the specific tensor at that point.

(psuedo-tensors, I suppose),

No. The metric tensor field is a field of tensors, not a field of pseudo-tensors.

one for every coordinate,

No. One tensor for every point. (In a 4-dimensional Lorentzian spacetime, every chart assigns 4 coordinates to every point within its chart, so writing "every coordinate" in the singular is wrong. More importantly, however, points are not the same as coordinates, which is obvious to everyone who understands that every spacetime can be described using quite a variety of coordinate systems (charts). But Mike Helland doesn't understand that, because he consistently conflates maps (i.e. charts) with territory (spacetime).)

which don't change the direction of the basis vectors, but effectively change their length to a bespoke value for each coordinate.

Mike Helland has provided abundant evidence that he doesn't understand the concept of a vector basis. The sentence quoted above is just another piece of that evidence.

It seems depending on who wrote the textbook/course, the wording on that may have several variations. But that seems to be the gist.

Yes, different authors may use varied terminology (e.g. "chart" versus "coordinate patch") and notations (e.g. sign conventions).

But all of that terminology and notation can be used correctly.

The author and sole proponent of Helland physics does not use any standard terminology or notation correctly. He gets just about everything wrong, and does so consistently.

 

[Mike Helland]

By plugging in the coordinates of a point, you obtain the specific tensor at that point.

Plugging in the coordinates?

I thought the metric tensor doesn't take coordinates as inputs.

So what's up with that?

Can you plug them "in", or not?

You can plug in, but don't call them inputs. Is that what you're saying?

 

[W.D.Clinger]

The author and sole proponent of Helland physics is so clueless that highlighting the relevant word in red bold italic didn't help.

 

[Mike Helland]

The author and sole proponent of Helland physics is so clueless that highlighting the relevant word in red bold italic didn't help.

So, the metric tensor doesn't take coordinates as inputs, but the metric tensor form does?

Moving on.

Let's say we input the coordinates of a point into the metric tensor form. We get back a tensor for that specific point. Which looks like a matrix of scalars, but its a tensor so there's more to it than that.

So how do those scalars relate to the tangent space?

* It seems to me every point has a tangent space defined for it.
* That each tangent space has a set of basis vectors for it.
* And the length of those basis vectors are defined by the scalars.

 

[Mike Helland]

The author and sole proponent of Helland physics is so clueless that highlighting the relevant word in red bold italic didn't help.

Do you have a resource that describes the distinction between a metric tensor and a metric tensor form?

Googling "metric tensor form" doesn't seem to come up with much, except when "form" is used as a verb.

 

[hecd2]

Do you have a resource that describes the distinction between a metric tensor and a metric tensor form?

Googling "metric tensor form" doesn't seem to come up with much, except when "form" is used as a verb.

Why don't you just learn physics and maths in a conventional way? If you started with high school maths and physics and worked your way through it systematically, you could understand all this stuff properly in less than five years. That would be much better than blundering around like a buffoon and learning nothing.