I My conclusions regarding "the laws of nature"

[davidge]

You may know I started a thread on this forum asking questions about the statement in General Relativity that "the laws of nature are the same in inertial frames". Guessing about the answers I got, I arrived in the following conclusion. I'd like to know whether these make sense & are correct or not.

On the Newton's laws:

1 - The validity of the second law implies the validity of the first law;
2 - The second law not being valid does not imply that the first law is not valid;
3 - The first law being valid or not does not imply the validity of the second law.

On (Maxwell's) electromagnetism & Newton's laws:

4 - The laws of electromagnetism have the same kind of invariance* than those of Newton.

Having these in mind, we could conclude:

Reason for invariance of light speed in inertial frames: Suppose there are two reference frames in which Newton's second law applies; from 1 above they are both inertial frames. Furthermore, as Newton's second law is invariant between them, so are Maxwell's equations (from 4). Conclusion: Speed of light is the same in both of them.

Newton's second law: Newton's second law is no longer valid in the relativistic theory because it doesn't remain invariant between two frames related by a Lorentz transformation, even that they are inertial frames (from 2).

I followed a response Dr. Greg gave me and arrived in the following invariant equation:

$$m \frac{d^2}{d \tau^2} x^\alpha = f^\alpha$$

where $m$ is the inertial mass and $\tau$ is the particle proper time (of a given particle, lol). This is invariant because it's the same in two inertial frames, in fact in all frames, since it's the time measured by a clock in the particle's reference frame. The equation above is not the same as Newton's second equation for two things; the first is that the time in the derivative is not the coordinate time and the other thing is that we are now dealing with four-vectors, which also means the force on the RHS is no longer the force on the corresponding RHS of Newton's second law. Note: I considered a Minkowskian framework on the equation above.

Main conclusion (because it motivated me to start the other thread): The statement that the laws of nature are the same in all inertial frames is not quite correct. Better would to say that they are valid as long as we keep ourselves in a theory where a invariance like the one in (*) exists. So I think one should specify in what theory they are going to talk about the laws of nature, because the invariance of laws of nature seems to depend up on the kind of invariance that equations obey.

There remains one question: how can we know whether a given reference frame is inertial? Apart from the obvious way of making an experiment to see whether Newton's first law works.* What I mean is that all vectors appearing in Maxwell's equations obey the same invariance rule than the one appearing on Newton's second law (the acceleration vector): $$ \vec{A} = (A_1 , A_2 , A_3) \\ (A_1)^2 + (A_2)^2 + (A_3)^2 = \text{constant} \\ $$ Obs. : usual metric above.

 

[dextercioby]

I think you are not prepared to understand the debunking of your misconceptions and perhaps PF is not the place for that.

 

[Nugatory]

The statement that the laws of nature are the same in all inertial frames is not quite correct.

You may have missed an important subtlety from the previous thread. Newton's second law is not exactly correct in any inertial frame so it is not a law of nature; therefore the principle of relativity should not be expected to apply. And because the second law is not exactly correct in any inertial frame, your chain of logic starting from "Suppose there are two reference frames in which Newton's second law applies..." is based on a false premise ("suppose something that doesn't exist exists") so does not lead to valid conclusions.

Of course the four-vector formulation $\vec{F}=d\vec{p}/dt$ in Minkowski space is exactly correct - but it is not equivalent to the three-vector formulation that we know as Newton's second law, it's a different law that makes different and more accurate predictions. This is in contrast with Maxwell's equations, where the three-vector $\vec{E}$, $\vec{B}$ formulation is exactly equivalent to the four-vector formulation written in terms of the faraday tensor.

 

[PeterDonis]

Guessing about the answers I got

You should not guess. You should not reason about this at all in ordinary language. To really understand this topic, you need to look at the actual math, and frame questions in terms of what the math is telling you. Even the most basic ordinary language words, like "principle of relativity" and "invariance", have shifting meanings in this context.

On the Newton's laws

Which ones? Newton's laws, as in Newtonian mechanics? Or Newton's laws, as in the modified versions we end up with when we take relativity into account and formulate the laws the way they need to be formulated in relativity?

If you are talking about Newton's laws, as in Newtonian mechanics, then your statement 1 is true, and your statements 2 and 3 make no sense. If you are talking about relativity, then your statement 1 makes no sense, your statement 2 is true, and your statement 3 makes no sense.

On (Maxwell's) electromagnetism & Newton's laws

Formulated in which framework? In the framework of Newtonian mechanics, where we use Galilean transformations to transform between inertial frames, Maxwell's Equations are not frame invariant; they are only valid in one particular inertial frame, and in any other inertial frame, they are invalid. Whereas Newton's laws are valid in all inertial frames.

But in the framework of relativity, where we use Lorentz transformations to transform between inertial frames, Maxwell's Equations are frame invariant (since they are Lorentz invariant), Newton's first law is still valid (and Lorentz invariant), Newton's second law in its usual form is not valid (since it isn't even formulated correctly to begin with), and there is a way to come up with an equation that "sort of looks like" Newton's second law and is valid (and Lorentz invariant).

None of the above scenarios make your statement 4 look correct.

Having these in mind, we could conclude

From here on your post is mostly speculation based on incorrect premises.

I followed a response Dr. Greg gave me and arrived in the following invariant equation:
$$
m \frac{d^2}{d \tau^2} x^\alpha = f^\alpha
$$
where mm is the inertial mass and $\tau$ is the particle proper time (of a given particle, lol). This is invariant because it's the same in two inertial frames, in fact in all frames, since it's the time measured by a clock in the particle's reference frame

This is correct, with the stipulation that it only applies to objects or systems whose invariant mass $m$ is constant in time. Not all objects and systems meet this requirement. For those that don't, there is a more general equation that is still valid, and invariant, and reduces to the one you wrote for the $m$ constant case.

However, it's also highly instructive to look at the physical interpretation of this equation. What is $x^\alpha$? What is $f^\alpha$? What do they mean in some particular concrete example?

The statement that the laws of nature are the same in all inertial frames is not quite correct

Sure it is. It just means that the laws of nature are Lorentz invariant. All of the known laws are.

how can we know whether a given reference frame is inertial? Apart from the obvious way of making an experiment to see whether Newton's first law works

That's the only way.

 

[Nugatory]

Y
* What I mean is that all vectors appearing in Maxwell's equations obey the same invariance rule than the one appearing on Newton's second law (the acceleration vector): $$ \vec{A} = (A_1 , A_2 , A_3) \\ (A_1)^2 + (A_2)^2 + (A_3)^2 = \text{constant} \\ $$ Obs. : usual metric above.

That's not invariance, that's the definition of the magnitude of a three-vector in Euclidean space. It's trivially true for for all three-vectors (including velocities and accelerations) in all reference frames. In general, however, the values of the individual components and the constant will be different in different reference frames, and that is as true of the electrical and magnetic fields as it is of velocities and accelerations.

 

[pervect]

Special relativity is not Newtonian mechanics, and F=ma just doesn't work in special relativity. Sorry. Bur rather than leave it at that, let's look at what does work. What does work is F = dp/dt, where p is momentum. So force is the rate of change of momentum with time. This works for both Newtonian mechanics and special relativity.

In Newtonian mechanics, the momentum is mass $\cdot$ velocity, so that p=mv, thus F = dp/dt = d(mv)/dt = m (dv/dt) = ma, as m is constant. (I suppose now is a good time to mention that by m, I mean the invariant mass, it's a concept of mass that's a property of an object and doesn't depend on the objects velocity).

In special relativity we have a different formula for momentum. Instead of p=mv, we write p =$\frac{mv}{\sqrt{1-(v/c)^2}}$

That's basically all there is to it. People get over-attached to F=ma (I can only guess why), and are reluctant to let go of it.

 

[davidge]

@pervect thanks for your contribution and hint.

Newton's second law is not exactly correct in any inertial frame so it is not a law of nature

Ah, ok. This is so "physically speaking", correct? Because when we go to the math..., see below

In the framework of Newtonian mechanics, where we use Galilean transformations to transform between inertial frames, Maxwell's Equations are not frame invariant

So, @Nugatory and @PeterDonis I think I should have been more clear on what I mean. So I'm going to add more details on what I meant in the opening post by invariance on electromagnetism and Newtonian mechanics.

The vector appearing in one side of Newton's equation is the (3-vector) acceleration: $d^2 \vec{x}/dt^2$. If the equation is to be keep form invariant, then we must end up in another "primed" coordinate system with $d^2 \vec{x'}/dt'^2$. As we know in Galilean transforms time $t'$ can be different from $t$ at most by a constant factor. So $dt' = dt$. As $\vec{x}$ is a three-vector which obeys (components here) $x^2 + y^2 + z^2 = x'^2 + y'^2 + z'^2$, then our acceleration vector is indeed a vector. What I said in the opening post about invariance of Maxwell's equations, I was talking about the separate equations, for example $\vec{\nabla} \times \vec{E} = -d \vec{B}/dt$. This is by construction a 3-vector whose components obey the same relation I wrote earlier on for the position vector. So it seems that the separate Maxwell's equations are invariant under Galilean transformations.

Now, the so called wave equation is not invariant.

I hope you understand what I'm trying to say.

 

[Dale]

So it seems that the separate Maxwell's equations are invariant under Galilean transformations

No, they are not. You are assuming that the Galilean transformation law for E and B are E=E' and B=B', but that would make the Lorentz force transform wrong. To save the Lorentz force law the Galilean transform for E and B are $E'=E +v \times B$ and $B'=B-v \times E$. Plug those in and you save the Lorentz force law, but then Maxwell's equations don't transform right.

 

[PeterDonis]

The vector appearing in one side of Newton's equation is the (3-vector) acceleration: $d^2 \vec{x}/dt^2$. If the equation is to be keep form invariant, then we must end up in another "primed" coordinate system with $d^2 \vec{x'}/dt'^2$.

That's not enough. You need the whole equation to take the same form, not just one side. So if in the original frame you have

$$
\vec{F} = m \frac{d^2 \vec{x}}{dt^2}
$$

then when you transform into the new frame, you must have

$$
\vec{F}' = m \frac{d^2 \vec{x}'}{dt'^2}
$$

As $\vec{x}$ is a three-vector which obeys (components here) $x^2 + y^2 + z^2 = x'^2 + y'^2 + z'^2$, then our acceleration vector is indeed a vector.

That's not the question you're trying to answer. The question you are trying to answer is whether the full equation of the second law takes the same form when we transform it from one inertial frame to another. Just showing that $\vec{x}$ "is a vector" isn't sufficient for that.

@Dale has already responded to your incorrect claim about Maxwell's Equations.

 

[PeterDonis]

As we know in Galilean transforms time $t'$ can be different from $t$ at most by a constant factor. So $dt' = dt$.

That's not how you transform a function of $t$. The quantity $d^2 \vec{x} / dt^2$ is not a fraction. It's a second derivative. In the original frame you have a function $\vec{x} (t)$. When you transform this function into the new frame, you get a different function $\vec{x}' (t)$. You need to show that these two functions have the same second derivative (assuming that you have already shown that the force $\vec{F}$ is unchanged by a Galilean transformation, i.e., that $\vec{F}' = \vec{F}$). What you have said so far is irrelevant to answering that question.

 

[davidge]

@Dale I got it. But why should we prefer Lorentz force being invariant over the Maxwell's equations? Why shoud'nt we keep Maxwell's equations invariant and let the Lorentz force not variant?

@PeterDonis ok, but since we are talking about a equation, that is, an equality, so I guess it suffices to show that one side is correct -- the other side will automatically be true for the equation to be a equation.

 

[Dale]

@Dale I got it. But why should we prefer Lorentz force being invariant over the Maxwell's equations? Why shoud'nt we keep Maxwell's equations invariant and let the Lorentz force not variant?

The Lorentz force is how E and B are defined usually. So I don't know how you could "opt out" of it.

 

[Herman Trivilino]

Of course the four-vector formulation $\vec{F}=d\vec{p}/dt$ in Minkowski space is exactly correct - but it is not equivalent to the three-vector formulation that we know as Newton's second law, it's a different law that makes different and more accurate predictions.

I was under the impression that $\vec{F}=d\vec{p}/dt$ (as a three-vector relation) is valid in special relativity. Is this not true?

 

[davidge]

The Lorentz force is how E and B are defined usually

If we find a definition of E and B other than that, so is it ok to keep Maxwell invariant instead of the Lorentz force?

 

[PeterDonis]

since we are talking about a equation, that is, an equality, so I guess it suffices to show that one side is correct -- the other side will automatically be true for the equation to be a equation

We are talking about taking an equation that is valid in one frame, and transforming both sides, and seeing if it is still valid in the new frame. To check this, you have to check both sides, because you transformed both sides; the fact that they were equal before the transform does not automatically guarantee that they will be equal after the transform.

 

[Dale]

If we find a definition of E and B other than that, so is it ok to keep Maxwell's invariant instead of the Lorentz force?

If you find one in the professional scientific literature that does that using the Galilean transform then it would be fine.

 

[SiennaTheGr8]

I was under the impression that $\vec{F}=d\vec{p}/dt$ (as a three-vector relation) is valid in special relativity. Is this not true?

It's just a definition.

 

[davidge]

The Lorentz force is how E and B are defined usually

that would make the Lorentz force transform wrong

I want just a little clarification on this.

The usual Lorentz force you talked about is a 3-vector equation. We have $\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$. Is your last statement quoted above equivalent to saying that under a Galilean transformation $\vec{F}$ does not transform such that $\vec{F}{}' = \vec{F}$?

I'm going to show a simple calculation

. The same should hold true for the vector $\vec{v} \times \vec{B}$. Therefore the Lorentz force should be invariant under Galilean transforms. Am I doing something wrong? OBS: I've made use of $d / dx$, etc because they transform as vectors. But I think I should have used $e_x, e_y, e_z$ instead.

 

[Umaxo]

We are talking about taking an equation that is valid in one frame, and transforming both sides, and seeing if it is still valid in the new frame. To check this, you have to check both sides, because you transformed both sides; the fact that they were equal before the transform does not automatically guarantee that they will be equal after the transform.

I would say it does quarantee, does it not? If you transform everything correctly, then the equation must still hold. Those are vector equations, so they are valid no matter the coordinates you are using. (I´d say it is true for general equation, not only vector equations, it is just that the transformation laws for those objects will be nontrivial)

However, in different coordinates the actual form of the equation will be different. And then, you are interested which set of coordinate transformations keeps the equation invariant in form - like in Newtonian mechanics those are galileo transformations.

 

[Ibix]

I would say it does quarantee, does it not? If you transform everything correctly, then the equation must still hold.

Yes, but only if you assume that the equations are consistent with your underlying assumptions. One of the reasons we ended up with relativity is that Maxwell's equations don't transform consistently under Galilean transforms. So I think one should be wary of "this half looks right, the other half must be".

 

[Umaxo]

Yes, but only if you assume that the equations are consistent with your underlying assumptions.

Which underlying assumptions are you referring to?

So I think one should be wary of "this half looks right, the other half must be".

I didnt try to say anything like that. I said, the equality holds, but not the form. Therefore both halves of the equation must be checked to see wheter the form remained same or not. If so, then the solutions to the equations are mathematicly the same and therefore the physics cannot distinguish between one frame and the other. If form is not the same, as it is with maxwell equations under galilean transformations, then solutions also don't look the same (in general) so physics can (in general) distinguish between those frames.

That is why first attempts after realizing that Maxwell equations are not invariant under galilean transformations was to find the special frame of reference or some kind of dynamical explanation instead of developing modified mechanics to keep principle of relativity (I believe einstein was in fact the first one who took later approach after 20 years of research in the area, and it is the reason why not even poincare nor lorentz could recognize new mechanics even though they have basicly everything ready?).

 

[PeterDonis]

If you transform everything correctly, then the equation must still hold.

Not unless you define "transform correctly" to make the equation still hold. For example, if you want Maxwell's Equations to still hold after being "transformed correctly", then you have to define "transform correctly" as "Lorentz transformation". But you can't just arbitrarily define what "transform correctly" means in general; your physical theory is supposed to tell you that. Newtonian mechanics says "transform correctly" is defined as "Galilean transformation", not "Lorentz transformation", so in Newtonian mechanics, Maxwell's Equations do not still hold after being transformed, even if you "transform correctly".

I said, the equality holds, but not the form

This is playing with words. If you transform Maxwell's Equations with a Galilean transformation, you don't end up with Maxwell's Equations in the new frame; you end up with different equations. Saying "well, the equality holds, but not the form" just obfuscates this obvious fact. Nobody cares about "equality" if the "form" changes; it's the "form" that contains the physics.

 

[davidge]

It's astonishing I'm obtaining that some of Maxwell's equations are invariant under Galilean transformations. I have done the calculations all the way here.

 

[PeterDonis]

It's astonishing I'm obtaining that some of Maxwell's equations are invariant under Galilean transformations. I have done the calculations all the way here.

Please show your work.

 

[Herman Trivilino]

It's just a definition.

I agree. Being a definition it's valid, by definition.

Looking back, I think I misunderstood the comment made by @Nugatory.

 

[davidge]

Please show your work.

I could do that. However, I think I used too many assumptions in my work. At a later time I found a page on web where the guy derives it in a way you and others here will probably accept as valid. Here is the link: https://physics.stackexchange.com/q...n-invariance-of-a-subset-of-maxwell-equations

 

[Dale]

Am I doing something wrong?

Yes, this approach is not showing what you want to show. What we are trying to show is how the whole equation transforms. So we want to show that when you transform $F=q(E+v\times B)$ you get $F'=q'(E'+v'\times B')$. Under the Galilean transform $v'=v+u$ and $F'=F$ from the definition of the Galilean transform and the Galilean invariance of Newton's 2nd law. I don't remember the reasoning why $q=q'$ but I am sure you can dig that up somewhere.

So consider $u=-v$, then $q'E'=F'=F=q(E+v\times B) \ne q E$. Similarly for B.

 

[PeterDonis]

At a later time I found a page on web where the guy derives it in a way you and others here will probably accept as valid.

No, actually, I don't. Perhaps it will help to write down the correct expressions for transforming derivatives with respect to the coordinates. For simplicity, I'll consider the case of one spatial dimension, so the Galilean transformation is $x' = x - vt$, $t' = t$, and the inverse transformation is $x = x' + v t'$, $t = t'$. Then we have

$$
\frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} + \frac{\partial t}{\partial x'} \frac{\partial}{\partial t}
$$

$$
\frac{\partial}{\partial t'} = \frac{\partial x}{\partial t'} \frac{\partial}{\partial x} + \frac{\partial t}{\partial t'} \frac{\partial}{\partial t}
$$

From the inverse transform equations, we obtain $\partial x / \partial x' = \partial t / \partial t' = 1$, $\partial x / \partial t' = v$, and $\partial t / \partial x' = 0$. This gives

$$
\frac{\partial}{\partial x'} = \frac{\partial}{\partial x}
$$

$$
\frac{\partial}{\partial t'} = v \frac{\partial}{\partial x} + \frac{\partial}{\partial t}
$$

So, contrary to the claim made in the link you gave, we do not have $\partial / \partial t = \partial / \partial t'$.

Furthermore, when transforming the fields $\vec{E}$ and $\vec{B}$, we have to remember that these are functions of the coordinates, and they will not, in general, be the same functions in both frames. For example, suppose we have an electric field that, in the unprimed frame, is that of a point charge at rest, i.e., $\vec{E} = Q / r^2$ and $\vec{B} = 0$. In the primed frame, this will be the field of a moving charge, so at a given spatial point, the distance from the charge will vary with time (so $\vec{E}'$ will now be a function of $t'$ as well as $r'$), and there will be a nonzero magnetic field $\vec{B}'$. So you can't just assume that $\vec{E}' = \vec{E}$ and $\vec{B}' = \vec{B}$.

 

[Umaxo]

Not unless you define "transform correctly" to make the equation still hold. For example, if you want Maxwell's Equations to still hold after being "transformed correctly", then you have to define "transform correctly" as "Lorentz transformation". But you can't just arbitrarily define what "transform correctly" means in general; your physical theory is supposed to tell you that. Newtonian mechanics says "transform correctly" is defined as "Galilean transformation", not "Lorentz transformation", so in Newtonian mechanics, Maxwell's Equations do not still hold after being transformed, even if you "transform correctly".

I´d say that equations says that certian mathematical object on the right side is the same as mathematical object on the left side. Thus if you transform it in whatever way, you are doing the same procedure on both sides, thus equality still holds. I meant "transform correctly" in this mathematical way. Does it make sense?

I was just curious because this is how i see equations, so when you said " the fact that they were equal before the transform does not automatically guarantee that they will be equal after the transform" it made me wondering.

This is playing with words. If you transform Maxwell's Equations with a Galilean transformation, you don't end up with Maxwell's Equations in the new frame; you end up with different equations. Saying "well, the equality holds, but not the form" just obfuscates this obvious fact. Nobody cares about "equality" if the "form" changes; it's the "form" that contains the physics.

I didnt say you end up with maxwell equations, just that you end up with equality.

 

[Dale]

@davidge

http://fisica.unipv.it/percorsi/pdf/jmll.pdf

It looks like my transformation was a little off

 

[vanhees71]

I want just a little clarification on this.

The usual Lorentz force you talked about is a 3-vector equation. We have $\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$. Is your last statement quoted above equivalent to saying that under a Galilean transformation $\vec{F}$ does not transform such that $\vec{F}{}' = \vec{F}$?

I'm going to show a simple calculation

View attachment 211363. The same should hold true for the vector $\vec{v} \times \vec{B}$. Therefore the Lorentz force should be invariant under Galilean transforms. Am I doing something wrong? OBS: I've made use of $d / dx$, etc because they transform as vectors. But I think I should have used $e_x, e_y, e_z$ instead.

Of course, if you do a non-relativistic calculation you cannot expect that it's giving the correct transformation laws. The most simple way is to use a covariant notation. For the mostion of a charged particle in an external electromagnetic field, given in terms of the Faraday tensor $F_{\mu \nu}$ the relativistic equations of motion read
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2} = \frac{q}{c} F^{\mu \nu}(x)\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}.$$
Here $\tau$ is the proper time of the particle. As it must be, both sides transform as contravariant vector components under Lorentz transformations.

For more details and also simple examples, see
http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

I don't see too much merit of studying non-relativistic electrodynamics (except in the usual cases, where the non-relativistic treatment of the matter particles is sufficient).

 

[PeterDonis]

I meant "transform correctly" in this mathematical way. Does it make sense?

It's still playing with words. See below.

I didnt say you end up with maxwell equations, just that you end up with equality.

In other words, you're ok with still calling it "equality" when the physics predicted by the "equality" is changed. That's not a useful definition of "equality" in a discussion about physics, particularly not when the topic under discussion is precisely the question of the invariance of the physics under transformations from one frame to another.

 

[davidge]

@PeterDonis , it's not said there that the two times $t'$ and $t$ are equal. I think you looked at the question, not the answer, because it's the questioner who says $t'=t$.

Yes, this approach is not showing what you want to show. What we are trying to show is how the whole equation transforms. So we want to show that when you transform $F=q(E+v\times B)$ you get $F'=q'(E'+v'\times B')$. Under the Galilean transform $v'=v+u$ and $F'=F$ from the definition of the Galilean transform and the Galilean invariance of Newton's 2nd law. I don't remember the reasoning why $q=q'$ but I am sure you can dig that up somewhere.

So consider $u=-v$, then $q'E'=F'=F=q(E+v\times B) \ne q E$. Similarly for B.

I see. As $q'=q$ we have to conclude that $E' \neq E$. But what about the link I posted just before your post? Maybe you want to look up the answer to the question that is posed there.

 

[PeterDonis]

it's not said there that the two times $t'$ and $t$ are equal

Nor is that what I was talking about. Go back and read my post again, carefully.

 

[vanhees71]

Yes, this approach is not showing what you want to show. What we are trying to show is how the whole equation transforms. So we want to show that when you transform $F=q(E+v\times B)$ you get $F'=q'(E'+v'\times B')$. Under the Galilean transform $v'=v+u$ and $F'=F$ from the definition of the Galilean transform and the Galilean invariance of Newton's 2nd law. I don't remember the reasoning why $q=q'$ but I am sure you can dig that up somewhere.

So consider $u=-v$, then $q'E'=F'=F=q(E+v\times B) \ne q E$. Similarly for B.

I can't say what you are doing with Galilei transformations, and it's way more complicated to consider non-relativistic electrodynamics than the full theory, because the literally natural realm of electromagnetism is relativistic physics. That's why historically relativity has been discovered by many people at the beginning of the 20th century (and finalized and fully understood by Einstein in 1905 and Minkowski in 1908) by thinking about electromagnetic phenomena.

So why is electric charge a scalar? Using the Maxwell equations alone (no need for closing the system by the mechanical equations for the matter thanks to gauge invariance), you get
$$\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$
You can write this in a covariant way as
$$\partial_{\mu} j^{\mu}=0,$$
and now using the 4D Gauss's integral theorem you find out that indeed the total charge
$$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \rho'=\text{const}.$$
For the detailed argument, see

http://th.physik.uni-frankfurt.de/~hees/publ/kolkata.pdf (pages 18+19)

 

[davidge]

So, contrary to the claim made in the link you gave, we do not have $\partial / \partial t = \partial / \partial t'$.

Sorry, but there's no such claim on there.
I ask you to please revisit the page and look at the most-liked answer.

 

[PeterDonis]

please revisit the page and look at the most-liked answer.

It would have been helpful to make it clear that that particular item on the page was what you were referring to. I'll take a look.

 

[PeterDonis]

I'll take a look.

After taking a look I'm not sure I agree with the most-liked answer's derivation either. The problem I see is the claim that the Lorentz force doesn't depend on acceleration, therefore $\vec{F}' = \vec{F}$ under Galilean transformation is true.

What is known to be true under Galilean transformation is that Newton's second law remains invariant, i.e., $m \vec{a}' = m \vec{a}$. (We're assuming that $m$ is invariant since we're talking about a single particle.) But the claim being made is that the Lorentz force formula holds in both frames, i.e., that $m \vec{a}' = q \left( \vec{E}' + \vec{v}' \times \vec{B}' \right)$ if $m \vec{a} = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$. I'm not sure this is correct; in any event, the argument given does not establish this claim.

The original post in the Stack Exchange thread links to a paper discussing a very different argument due to Feynman:

https://arxiv.org/pdf/hep-ph/0106235.pdf

I'm looking at that now.

 

[Nugatory]

But the claim being made is that the Lorentz force formula holds in both frames, i.e., that $m \vec{a}' = q \left( \vec{E}' + \vec{v}' \times \vec{B}' \right)$ if $m \vec{a} = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$. I'm not sure this is correct

This claim is equivalent to the claim that an accelerometer attached to our test particle will read the same in both frames, is it not?

 

[PeterDonis]

This claim is equivalent to the claim that an accelerometer attached to our test particle will read the same in both frames, is it not?

No. That claim is the claim that $\vec{a}' = \vec{a}$.

The claim I'm questioning is the claim that the relationship between $\vec{a}$ and the fields $\vec{E}$ and $\vec{B}$ and the velocity $\vec{v}$, which is what the Lorentz Force equation expresses, is the same in both frames. That claim is a claim about the relationship between accelerometer measurements, velocity measurements, and field strength measurements. It's not a claim about accelerometer measurements alone.

 

[PeterDonis]

I'm looking at that now.

Having looked, I don't think Feynman's derivation is relevant here, since it assumes that position and velocity do not commute (equation 2 of the paper). While this is very interesting from the standpoint of relating non-relativistic electrodynamics to quantum mechanics, it is irrelevant to this discussion since we are comparing classical (non-quantum) Newtonian mechanics and special relativity.

 

[Dale]

I can't say what you are doing with Galilei transformations, and it's way more complicated to consider non-relativistic electrodynamics than the full theory, because the literally natural realm of electromagnetism is relativistic physics

I agree 100%. I am very likely making mistakes with the Galilei transformations since I see little value in spending a lot of effort learning how to do a wrong thing correctly.

 

[Dale]

Maybe you want to look up the answer to the question that is posed there.

I did read it, but an incomplete subset of Maxwell's equations is of fairly limited applicability. EM requires all of them and also the Lorentz force law to be complete.

Maybe you want to read the more complete treatment: http://fisica.unipv.it/percorsi/pdf/jmll.pdf

 

[davidge]

@PeterDonis, I think your disconfort with the invariance of the Lorentz force, i.e. $F' = F$ can be resolved by noticing that for a particle $q(E + v \times B) = d (m v) / dt$, i.e. the RHS is Newton's second law, which is invariant under Galilean transformations. So we must have invariance of the Lorentz force too. I think this is the reasoning of the author of that post.

Maybe you want to read the more complete treatment: http://fisica.unipv.it/percorsi/pdf/jmll.pdf

I will read it when I have the time.

 

[PeterDonis]

the RHS is Newton's second law, which is invariant under Galilean transformations

Sorry, you are still ignoring the actual point I'm making. Go read my post #38 in response to @Nugatory.

So we must have invariance of the Lorentz force too

This reasoning is invalid. Again, go read post #38.

 

[vanhees71]

No. That claim is the claim that $\vec{a}' = \vec{a}$.

The claim I'm questioning is the claim that the relationship between $\vec{a}$ and the fields $\vec{E}$ and $\vec{B}$ and the velocity $\vec{v}$, which is what the Lorentz Force equation expresses, is the same in both frames. That claim is a claim about the relationship between accelerometer measurements, velocity measurements, and field strength measurements. It's not a claim about accelerometer measurements alone.

As I said before, it's much simpler to study the exact special-relativistic equation. There's no merit in the non-relativistic limit here at all. Of course you can get easily the non-relativistic limit in the usual way:

The covariant equation reads
$$m \ddot{x}^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu},$$
where dots denote derivatives wrt. proper time, $m$ the invariant mass of the particle (scalar), $q$ the charge of the particle (scalar), and $F^{\mu \nu}$ the components of the Faraday tensor of the given field.

First of all, not all four EoMs are independent since, as it must be the equation is compatible with the constraint, following from the definition of proper time
$$\dot{x}_{\mu} \dot{x}^{\nu}=c^2 =\text{const}.$$
Thus it is sufficient to consider the spatial components only, and that reads very familiar when expressed in the (1+3)-form of the equation in the given reference frame:
$$m \ddot{\vec{x}}=q \left (\vec{E}+\frac{\dot{\vec{x}}}{c} \times \vec{B} \right).$$
Although this looks like the non-relativistic equation, it's not, because the dot denotes a derivative wrt. to proper time, $\tau$, rather than "coordinate time" $t$. Nevertheless this equation is form-invariant unter Lorentz transformations since it's derived from a equation only involving tensor components.

Then, in any frame, where the non-relativistic approximation is valid, i.e., in a frame, where (during the considered time!) at all times
$$\left |\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \right| \ll c.$$
Then you can approximate
$$\dot{f}=\frac{\mathrm{d} f}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} = \mathrm{d}_t f \frac{1}{\sqrt{1-(\mathrm{d}_t \vec{x})^2/c^2}} = \mathrm{d}_t f [1+\mathcal{O}(v^2/c^2)]$$
with $v=|\mathrm{d}_t \vec{x}|$.

Thus, using the correct relativistic equation and the correct transformation between frames (Lorentz transformation) leads to the conclusion that also the non-relativistic approximation is forminvariant in all frame it applies. I don't know whether the somewhat cumbersome calculation with Galilei transformations here really helps. In any case you have to take the corresponding limit of the transformation law for the field components too. The above covariant approach, however, suggests that this should lead to the same conclusion if done properly (i.e., taking consistent approximations in powers of $v/c$ on both sides of the equation).

 

[vanhees71]

I agree 100%. I am very likely making mistakes with the Galilei transformations since I see little value in spending a lot of effort learning how to do a wrong thing correctly.

This is so, because the non-relativistic limit of E&M is not as trivial as it seems. There are even two different limits (an "electric" and a "magnetic" one), depending on the actual physical situation. I think the relevant work by LeBellac et al has been already cited in this thread.

 

[haushofer]

This is so, because the non-relativistic limit of E&M is not as trivial as it seems. There are even two different limits (an "electric" and a "magnetic" one), depending on the actual physical situation. I think the relevant work by LeBellac et al has been already cited in this thread.

Levy-leblond has written a lot of nice papers about this and other issues in non-relativistic physics already in the '60s. One non-relativistic limit can be regarded as the c --> oo limit in which the replacement current vanishes. From my memory this replacement current term is exactly the term which spoils Galilei-covariance and enforces Poincaré-invariance on the Maxwell equations.

In my thesis I worked out a bit of the spacetime-covariant form of the Galilei-transformations and its metrical structure. It's textbook material, but maybe some finds it usefull (page 36,37):

http://www.rug.nl/research/portal/p...ed(fb063f36-42dc-4529-a070-9c801238689a).html

 

[vanhees71]

What I had in mind was the classical paper on the subject,

M. LeBellac, J.-M. Levi-Leblond, Galilean Electromagnetism, Nuovo. Cim. 14, 217 (1973)

Also see the references in

https://en.wikipedia.org/wiki/Galilean_electromagnetism

particularly Ref. 1.

 

[robphy]

What I had in mind was the classical paper on the subject,

M. LeBellac, J.-M. Levi-Leblond, Galilean Electromagnetism, Nuovo. Cim. 14, 217 (1973)

Also see the references in

https://en.wikipedia.org/wiki/Galilean_electromagnetism

particularly Ref. 1.

In #41, @Dale posted a link to a scan of that document (the link is also in the Wikipedia entry).

I have had a long interest (my posts from 2005 2006) in that paper due to this article

• "If Maxwell had worked between Ampère and Faraday: An historical fable with a pedagogical moral"
  Max Jammer, John Stachel , American Journal of Physics -- January 1980 -- Volume 48, Issue 1, pp. 5-7, http://dx.doi.org/10.1119/1.12239

and how it relates to Cayley-Klein geometry

• "Possible Kinematics"
  Henri Bacry, Jean‐Marc Lévy‐Leblond, Journal of Mathematical Physics 9, 1605 (1968); http://dx.doi.org/10.1063/1.1664490
• "A simple non-Euclidean geometry and its physical basis : an elementary account of Galilean geometry and the Galilean principle of relativity"
  I.M. Yaglom (1969,1979) https://archive.org/details/ASimpleNon-euclideanGeometryAndItsPhysicalBasis

which I regard as a way to formalize the classical Galilean limits and the geometrical analogies between spacetime and the Cayley-Klein Geometries.

Note: one should not confuse "Galilean Electromagnetism"
with a journal with a similar name (which was raised in a old 2008 PF thread: Galilean Electrodynamics? )