What Medium Does Light Travel Through in Space?

[Sagittarius A-Star]

The EM-wave function follows from Maxwells equations.

Derivation of the EM wave function (a linear, second-order partial differential equation) from Maxwells equations in (3+1) notation, Gaussian CGS units (see posting #107)

Write the $curl$ of Faradays law:
$\nabla \times (\nabla \times \mathbf{E}) = \nabla \times \left( - \frac{\partial \mathbf{B}}{\partial (ct)} \right)$
Apply the vector identity $\nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A}$ on the LHS and switch order of partial and time derivatives on the RHS:
$\nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} = -\frac{\partial}{\partial (ct)} (\nabla \times \mathbf{B})$
Substitute Gauss's law ($\nabla \cdot \mathbf{E} = 0$) and Ampere's law for ($\nabla \times \mathbf{B}$):
$$\nabla^2 \mathbf{E} - \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} = 0$$

 

[Dale]

Even if a dimension parameter such as the speed of light is numerically equal to 1 in a specific set of units one should strictly speaking still provide the units. So the statement c=1 would imply that c is dimensionless, which is not correct

The dimension of a quantity also depends on the system of units. The dimension is a convention adopted by the unit system. This is the big lesson of Gaussian and other cgs units. They are not just different sizes of the same quantities, but dimensionally different quantities.

When I say $c=1$, I mean it literally. I mean $c$ is a dimensionless quantity with a value of 1. It is like the radian in that sense.

Geometrized units are an example of this type of unit system.

The strictly correct expression should be something like c=1[units].

No. $c=1$ is what I intended. Although there are natural unit systems where length and time have different dimensions, when I use natural units I typically choose them to have the same dimension.

If I intend $c$ to be dimensionful then I will write it with units, eg $c=1 \mathrm{\ lightyear/year}$, specifically to indicate that I am considering it to be a dimensionful quantity.

Strictly speaking, the reason why such quantities "disappear" is because they are absorbed into other quantities. A change in units does not allow one to remove a quantity from an equation.

This is incorrect. In Gaussian units $\epsilon_0$ simply doesn’t exist. You aren’t removing it, it was never there to begin with. Coulomb’s law is already dimensionally consistent without it, so it never gets introduced in the first place.

Alternatively, you can think of the Gaussian $\epsilon_0$ as being a dimensionless 1. But that is just “training wheels” for people that cannot quite let go of SI units.

another quantity must be change by the inverse of that factor (if they are on the same level above or below the line) to ensure that dimensionless quantity remains the same

Yes. Or some other power of the factor.

 

[GRQFT]

Yes. The RHS of the wave function contains a 2nd derivative for $ct$. In SR, $t$ is multiplied with $c$ as conversion factor.

You will see SR in Maxwells equations, if you write them in 4-tensor notation.

Greek indices 1...4 stand for the spacetime coordinates:
$x^\mu = (x^1, x^2, x^3, x^4) = (x, y, z, ct)$

The EM field in Gaussian CGS units and convention (+++-):
$F^{\mu\nu} = \begin{pmatrix} 0 & B_z & -B_y & -E_x \\ -B_z & 0 & B_x & -E_y \\ B_y & -B_x & 0 & -E_z \\ E_x & E_y & E_z & 0 \end{pmatrix}$
The dual field tensor:
$\tilde{F}^{\mu\nu} = \begin{pmatrix} 0 & -E_z & E_y & -B_x \\ E_z & 0 & -E_x & -B_y \\ -E_y & E_x & 0 & -B_z \\ B_x & B_y & B_z & 0 \end{pmatrix}$
The 4-current:
$J^\mu = (J_x, J_y, J_z, c\rho)$

Maxwells 4 inhomogeneous equations (you get each by setting $\mu = 1...4$):
$\partial_\nu F^{\mu\nu} = \frac{4\pi}{c} J^\mu$

Maxwells 4 homogeneous equations by using the dual field tensor (you get each by setting $\mu = 1...4$):
$\partial_\nu \tilde{F}^{\mu\nu} = 0$ (no magnetic monopole current)

The EM-wave function follows from Maxwells equations.

Thanks so much for taking the time to write all that out!

So, to make sure I understand, you're saying that because Maxwells equations are Lorentz invariant (like SR which is foundationally Lorentz invariant), that's how we can see SR in Maxwells equations?

And as a bonus question, Maxwells equations were always Lorentz invariant (in fact I think they may have been the first to have that property), but writing them in 4-tensor notation simply makes that relationship more clear?

 

[GRQFT]

The EM field in Gaussian CGS units and convention (+++-):
$F^{\mu\nu} = \begin{pmatrix} 0 & B_z & -B_y & -E_x \\ -B_z & 0 & B_x & -E_y \\ B_y & -B_x & 0 & -E_z \\ E_x & E_y & E_z & 0 \end{pmatrix}$
The dual field tensor:
$\tilde{F}^{\mu\nu} = \begin{pmatrix} 0 & -E_z & E_y & -B_x \\ E_z & 0 & -E_x & -B_y \\ -E_y & E_x & 0 & -B_z \\ B_x & B_y & B_z & 0 \end{pmatrix}$

Looking at what you wrote above makes me think that the notion that the magnetic field is really just an electric field viewed in the wrong frame of reference is actually false (I've read/heard this several times).

Judging from the two matrices above, it seems that they are quite equal; i.e., neither one is more 'fundamental' than the other. Is that correct?

If so, does that mean we could use either tensor above and get the same answers? That is, we have two equivalent options such that we can choose whichever one yields the simplest calculations?

 

[renormalize]

If so, does that mean we could use either tensor above and get the same answers? That is, we have two equivalent options such that we can choose whichever one yields the simplest calculations?

Yes, both tensors contain the same information (6 independent components) since one is just a rearrangment of the other. It's exactly analogous to the cross product $\mathbf{c}=\mathbf{a}\times\mathbf{b}$ of two vectors in 3D. Using tensor notation, the cross product is $c_{k}=\varepsilon_{ijk}a_{i}b_{j}\,$, but you can just as well work with the antisymmetric tensor $a_{ij}=\left(a_{i}b_{j}-b_{i}a_{j}\right)/2$ that's dual to $c_{k}$. Both contain the same 3 independent quantities.

 

[Matterwave]

Judging from the two matrices above, it seems that they are quite equal; i.e., neither one is more 'fundamental' than the other. Is that correct?

Basically, but with the important caveat that there are no magnetic monopoles. See the rest of the quoted message.

 

[GRQFT]

Yes, both tensors contain the same information (6 independent components) since one is just a rearrangment of the other. It's exactly analogous to the cross product $\mathbf{c}=\mathbf{a}\times\mathbf{b}$ of two vectors in 3D. Using tensor notation, the cross product is $c_{k}=\varepsilon_{ijk}a_{i}b_{j}\,$, but you can just as well work with the antisymmetric tensor $a_{ij}=\left(a_{i}b_{j}-b_{i}a_{j}\right)/2$ that's dual to $c_{k}$. Both contain the same 3 independent quantities.

When you say "...that's dual to $c_{k}$", are you talking about the Hodge dual?

 

[renormalize]

When you say "...that's dual to $c_{k}$", are you talking about the Hodge dual?

Yes, which in tensor notation in D-dimensions, just means to fully contract any given rank-k tensor with the D-dimensional Levi-Civita pseudo-tensor to get the rank (D-k) dual tensor. So in 3D, the dual of the rank-1 cross-product becomes a rank 3-1 = 2 antisymmetric tensor, and in 4D the dual of the rank-2 antisymmetric Maxwell field tensor becomes the rank 4-2 = 2 antisymmetric dual-field tensor.

 

[Matterwave]

just means to fully contract any given rank-k tensor with the D-dimensional Levi-Civita pseudo-tensor to get the rank (D-k) dual tensor.

Remember to use the metric to keep the indices in the right place. In GR this can bite. In SR this leads to some negative signs in the components.

 

[GRQFT]

Yes, which in tensor notation in D-dimensions, just means to fully contract any given rank-k tensor with the D-dimensional Levi-Civita pseudo-tensor to get the rank (D-k) dual tensor. So in 3D, the dual of the rank-1 cross-product becomes a rank 3-1 = 2 antisymmetric tensor, and in 4D the dual of the rank-2 antisymmetric Maxwell field tensor becomes the rank 4-2 = 2 antisymmetric dual-field tensor.

Okay, thanks.

There's clearly a lot that I need to learn about tensors! I was thinking of buying a copy of Sean Carroll's Spacetime and Geometry and reading the relevant sections. Would that be a good choice, or is there something better?

 

[Matterwave]

Okay, thanks.

There's clearly a lot that I need to learn about tensors! I was thinking of buying a copy of Sean Carroll's Spacetime and Geometry and reading the relevant sections. Would that be a good choice, or is there something better?

Do you want more of a solid mathematical foundation in differential geometry? Or just what's needed to do relativity? Or something in between? What's your current mathematical level?

It might be nice to start a separate thread on this question. Textbook suggestions have their own forum I think. :)

 

[GRQFT]

Do you want more of a solid mathematical foundation in differential geometry? Or just what's needed to do relativity? Or something in between?

I'm interested in learning enough to understand relativity (and QFT).

What's your current mathematical level?

I studied physics in college (decades ago), so I guess I'm at a level where I I know enough to ask about stuff like the Hodge dual, but not really understand it. Thus far I've just been googling the terms I don't know, but that's clearly not an optimal strategy (hence my desire for a textbook recommendation).

It might be nice to start a separate thread on this question. Textbook suggestions have their own forum I think. :)

Okay, I didn't know that. I'll ask there. Thanks again for all your help!

EDIT: I was going to start a new topic, but the question has already been posted several times yielding many useful suggestions. What a great forum!

 

[bhobba]

You will see SR in Maxwells equations, if you write them in 4-tensor notation.

Even more elegantly:

https://www.vttoth.com/CMS/physics-notes/123-the-electromagnetic-field-tensor

Thanks
Bill

 

[bhobba]

Because of the choice of length that we call 'meter'.

Nice exercise relating it back to the purely POR derivation of the lorentz transformations.

Thanks
Bill

 

[GRQFT]

Even more elegantly:

https://www.vttoth.com/CMS/physics-notes/123-the-electromagnetic-field-tensor

Thanks
Bill

Excellent!

I just learned that:

1. "...the theory of electromagnetism is really just the geometric theory of an arbitrary smooth vector field."

2. "...in the case of the electromagnetic field, both the equations used to define E and B and the field equations will automatically be satisfied in curved spacetime, under the conditions of general relativity."

3. "I have often seen E and B described as "real" physical quantities, as opposed to A, which is not "real" because it is undetermined to the extent that you can add ∇f to it. But what we see here suggests to me that E and B are far less "real" than A! (Arguably, though, the electromagnetic field tensor F is real in the sense that it is a "proper" tensorial quantity.)"

Item #3 seems a bit controversial to me; maybe I just need some time to let it sink in.

 

[bhobba]

Excellent!

Things get even more interesting when you learn about two important, though not often discussed, theorems of Wigner - the no-interaction theorem (this means fields must exist) and that all fields must be tensors.

Thanks
Bill

 

[bhobba]

Item #3 seems a bit controversial to me; maybe I just need some time to let it sink in.

Well, that depends on what you mean by real. If only philosophers would agree on that, it may be answerable, but since they do not, your guess is as good as mine.

Thanks
Bill

 

[GRQFT]

Things get even more interesting when you learn about two important, though not often discussed, theorems of Wigner - the no-interaction theorem (this means fields must exist) and that all fields must be tensors.

Thanks
Bill

Thanks. I'll look into those...

The size of my Zotero physics collection has grown considerably since I joined PF!

 

[Matterwave]

3. "I have often seen E and B described as "real" physical quantities, as opposed to A, which is not "real" because it is undetermined to the extent that you can add ∇f to it. But what we see here suggests to me that E and B are far less "real" than A! (Arguably, though, the electromagnetic field tensor F is real in the sense that it is a "proper" tensorial quantity.)"

I would be careful with the word "real" here and also careful how you use your symbols.

As 3-vectors, $\vec{E}$ and $\vec{B}$ are not individually proper tensorial objects when viewed from the perspective of 4-D spacetime of SR or GR. Only their combination in $F^{\mu\nu}$ is a proper tensor.

But this is quite analogous to what happens to $\phi$ and $\vec{A}$ in that they get combined into a 4-vector $A^\mu = (\phi, \vec{A})$ when we move to the relativistic view. From this perspective, both $A^\mu$ and $F^{\mu\nu}$ are proper tensors, while the 3-D versions $\vec{E}$, $\vec{B}$, $\phi$ and $\vec{A}$ are not.

So question, when you say "as opposed to A" do you mean the 4-vector $A^\mu$ or the 3-vector $\vec{A}$? Make sure you don't silently confuse what you are referring to.

It is usual (but not necessary) that in the purely classical view of the world, we take the electromagnetic field (as described properly by $F^{\mu\nu}$) to be the real physical quantity. Since $A^\mu$ exhibits indeterminacy due to gauge transformations it seems somewhat unnatural to ascribe physical reality to it. However, importantly, in the quantum mechanical view, this picture is complicated by the Aharonov-Bohm effect. Discussion of the Aharonov-Bohm effect would take this thread off-course so I will just mention it without giving a full description.

 

[flippiefanus]

When I say c=1, I mean it literally. I mean c is a dimensionless quantity with a value of 1. It is like the radian in that sense.

Would you agree that what it actually implies is that $c$ is being absorbed into the time coordinate?

 

[flippiefanus]

In Gaussian units ϵ0 simply doesn’t exist. You aren’t removing it, it was never there to begin with. Coulomb’s law is already dimensionally consistent without it, so it never gets introduced in the first place.

When comparing systems of equations in different systems of units, one should be able to trace such quantities to understand why they don't exist in one of the systems of equations. I have not done this exercise, but I still think that such a comparison will reveal that the quantity disappeared because it is absorbed into another quantity.

 

[Dale]

Would you agree that what it actually implies is that $c$ is being absorbed into the time coordinate?

No. If $c$ is a dimensionless 1 then you could equally say that it was absorbed into the space coordinate. Or that its cube was absorbed into the time coordinate and its fourth root was absorbed into the space coordinate. Far easier to just say it just doesn’t exist in those units. There is no reason for it to be absorbed anywhere and inherent ambiguity if you try to assert that it is absorbed somewhere.

one should be able to trace such quantities to understand why they don't exist in one of the systems of equations

Indeed. Just find the reason why it does exist in the other. The absence of that reason is the reason that it doesn’t exist in the one.

For instance, $\epsilon_0$ exists in SI units to make Coulomb’s law dimensionally consistent. That is why it does exist in SI. In Gaussian units Coulomb’s law is naturally dimensionally consistent. That is why it doesn’t exist in Gaussian units.

You can do a similar exercise with every dimensionful constant that is present in one system of units and not in another.

 

[renormalize]

Would you agree that what it actually implies is that $c$ is being absorbed into the time coordinate?

I would say instead that $c$ simply disappears by choosing uniform units, as I tried to explain in post #118.
Please indulge me by pondering an exactly analogous situation in 3D that persists to this day in US sea navigation. American maritime vessels measure distance traveled in nautical miles (NM), but the depth below them is given in fathoms (ftm). The relation between these quantities is $1\,\text{NM}=\mathfrak{c}\,\text{ftm}$ where the "mysterious" conversion factor $\mathfrak{c}$ is $1012.69$. But the rest of the world, not being wedded to imperial tradition, more rationally measures height, width and depth all in the same uniform units, namely meters. The rest of the world thinks that the "conversion" between transverse-distance and longitudinal-depth is simply unity (if they think about it all). So what happened to the "dimensionful" constant $\mathfrak{c}$? You can say it was absorbed into the depth coordinate, whereas I claim it disappeared by making the natural choice of uniform units for height, width, and depth. But in either view, the result is consistent with saying $\mathfrak{c}=1\,$, and is dimensionless, in uniform units.
And that same reasoning applies to $c$ in the unified theory of Minkowski spacetime.

 

[Sagittarius A-Star]

So, to make sure I understand, you're saying that because Maxwells equations are Lorentz invariant (like SR which is foundationally Lorentz invariant), that's how we can see SR in Maxwells equations?

That's one aspect. Another is, that 4D-notation makes the relativistic structure of Maxwells equations obvious.
This is also valid for the Lorentz force, which is in Gaussian units:
$\frac{dp^\mu}{d\tau} = \frac{q}{c}F^\mu{}_{\nu}u^\nu$. The EM field is here a tensorial coefficient between the 4-velocity and the 4-force per charge.

And as a bonus question, Maxwells equations were always Lorentz invariant (in fact I think they may have been the first to have that property), but writing them in 4-tensor notation simply makes that relationship more clear?

Yes, it makes that relationship more clear because tensor equations are covariant.

 

[PeroK]

We could phase the question like this.

Human beings are on the scale of 1-2 metres. And the human heart beats about once per second.

With these definitions of metre and the second, could the speed of light theoretically be, say, 100 m/s?

My understanding is that we would have to vary the fine structure constant. Otherwise, the speed of light has a somewhat fixed relationship to "terrestrial" speeds.

 

[Ibix]

I will state that alot of this is above my head but I have often wondered the same thing about the speed of light. Why is it what it is and not something else? I have watched threads that discuss this and am always disappointed to see the answer of "because the meter is the length that it is", or similar non answers. I came to the conclusion that it is what it is because we always measure it this way and we don't have a better answer. That at times has translated into: "because we always measure it this way and I don't feel comfortable saying we don't know why". It's nice to see someone say: "we don't know".

I think you haven't quite caught the point here.

We know why $c$ has the value it has - because when we picked some artifact and called it 1m long and picked some fraction of the day and called that 1s, we defined the value of $c$ (edit: in the modern SI this is switched around, but the defined $c$ is so defined to be consistent with the original artifact definition). Then, because $\alpha=e^2(4\pi\varepsilon_0\hbar c)^{-1}$, we know that you cannot change $c$ without also changing at least one of $\alpha$, $e$, $\varepsilon_0$ and $\hbar$. And it turns out that if you hold $\alpha$ constant, any change is nothing more than a unit redefinition. Dale gave an example of a unit-free measurement upthread:

For example, lay a fixed number of copper atoms in a line and count the number of cesium hyperfine transitions it takes for light to cross, reflect, and cross back.

The result of that is a bare number, so is immune to unit changes. And, as Dale said, the result does not change if you change $c$ and any combination of $\hbar$, $e$ and $\varepsilon_0$, because the latter leads to countervailing changes in the size of atoms that mean that if you (e.g.) make light faster you must also make atoms larger - so now light travels twice as many rod lengths in the same time, but your rod is only half as many atoms long (i.e. you sawed your meter rule in half and called both halves a meter, which doubled the numerical value of $c$ and the numerical value of the diameter of atoms).

That bare number result only changes if you change $\alpha$. That means that it's $\alpha$ that controls why $c$ is so large in convenient-for-humans units.

So we know quite a lot about why $c$ has the value it has and the implications of changing it. But the changes are only physically meaningful if they are accompanied by a change in $\alpha$, and we have no idea why that has the value it has.

 

[Dale]

We could phase the question like this.

Human beings are on the scale of 1-2 metres. And the human heart beats about once per second.

With these definitions of metre and the second, could the speed of light theoretically be, say, 100 m/s?

My understanding is that we would have to vary the fine structure constant. Otherwise, the speed of light has a somewhat fixed relationship to "terrestrial" speeds.

Yes. You are emphasizing the dimensionless physical comparisons here. Not the value in some units.

This is indeed governed by the fine structure constant.

 

[Sagittarius A-Star]

Looking at what you wrote above makes me think that the notion that the magnetic field is really just an electric field viewed in the wrong frame of reference is actually false (I've read/heard this several times).

Judging from the two matrices above, it seems that they are quite equal; i.e., neither one is more 'fundamental' than the other. Is that correct?

That's correct. However, for certain scenarios the electric force on a test-charge in it's rest-frame can transform into a pure magnetic force in a specific other frame:
https://physics.weber.edu/schroeder/mrr/MRRtalk.html

If so, does that mean we could use either tensor above and get the same answers? That is, we have two equivalent options such that we can choose whichever one yields the simplest calculations?

The first tensor in posting #119 is the EM field, the second is the dual of it.
$\widetilde{F}^{\mu\nu} \equiv \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$

We can for example write Maxwells homogeneous equation with either version in the following way. A magnetic current $J_{\mathrm m}^{\nu}$ does not exist.

Using the EM field (with exterior derivative):
$\partial_\lambda F_{\mu\nu} +\partial_\mu F_{\nu\lambda} +\partial_\nu F_{\lambda\mu} = \frac{4\pi}{c}\, \widetilde{J}_{{\rm m}\,\lambda\mu\nu} = 0$.

Using the dual EM field (with interior derivative):
$\partial_\mu \widetilde{F}^{\mu\nu} = \frac{4\pi}{c}\,J_{\mathrm m}^{\nu} =0$.

 

[Dale]

one should be able to trace such quantities to understand why they don't exist in one of the systems of equations

As an exercise for you, trace why the constant $k$ in Newton’s 2nd law, $\vec F_{net}=km \vec a$, doesn’t exist in SI units.

Also, if you want to assert that $k$ does exist in SI, but is absorbed somewhere, then where is it absorbed and how do you know it was absorbed there and not somewhere else?

 

[GRQFT]

I would be careful with the word "real" here and also careful how you use your symbols.

To be clear, that was not me talking about the 'realness' of E & B, but rather Viktor T. Toth's claim on this webpage (see last paragraph).

So question, when you say "as opposed to A" do you mean the 4-vector $A^\mu$ or the 3-vector $\vec{A}$? Make sure you don't silently confuse what you are referring to.

I was referring to the 3-vector $\vec{A}$ (which I now realize is problematic given what you said about the 3-D classical view of $\vec{E}$, $\vec{B}$ vs the 4-vector relativistic view).

However, importantly, in the quantum mechanical view, this picture is complicated by the Aharonov-Bohm effect. Discussion of the Aharonov-Bohm effect would take this thread off-course so I will just mention it without giving a full description.

Yes, a wise choice given how controversial that subject is (and the fact that Aharonov himself has changed his interpretation of the effect [which he discusses in this video]).