Principle of relativity: active vs passive point of view

[cianfa72]

TL;DR Summary

Galilean principle of relativity from the active vs passive point of view

Hi,
starting from this thread I'm a bit confused about the content of the principle of relativity from a mathematical point of view.

Basically the "Galilean principle of Relativity" puts requirements on the transformation laws between Inertial Frame of Reference (IFR); thus they have to preserve the form of all the per-frame laws.

From a transformation perspective, I'm aware of there exist two interpretation: active (Alibi) and passive (Alias). Take a physical law mathematically described as $F(x, \dot x) = 0$ in a IFR and apply it to an experimental setup (call it "first experiment copy") obtaining the solution $x(t)$.

In the active point of view basically we consider basically a "second copy" of the experimental setup and by transformation calculate the coordinate values for that "second copy" in the IFR. Then entering those coordinate values in $F(x, \dot x) = 0$ we claim that they solve the equation if only if the coordinate values for the "first copy" do.

Consider now the other point of view (passive): this time there exist just "one copy" of the experimental setup described using two different coordinate systems (first IFR and transformed one coordinate systems).

Are the two point of view actually equivalent ?

Thanks in advance.

 

[haushofer]

Yes, on the level of equations of motion (!) these views are fully equivalent.

Of course, "ontologically" there is a difference between changing your own frame of reference, or moving around the object of interest. But what is invariant, is the relative orientation of that object with respect to your frame. And that orientation is described by your coordinates, which enter the equation of motion.

 

[haushofer]

Maybe these notes of mine also help:

https://www.physicsforums.com/threa...ssive-and-active-interpretations-etc.1000157/

 

[cianfa72]

Of course, "ontologically" there is a difference between changing your own frame of reference, or moving around the object of interest. But what is invariant, is the relative orientation of that object with respect to your frame. And that orientation is described by your coordinates, which enter the equation of motion.

Thus mathematically we can state the Principle of Relativity (both in Newton mechanics and SR) in that two equivalent forms.
Then in Newton mechanics, starting from $$F = m \frac {dx^2} {dt^2} $$ supposed to be valid in a IRF, the requirement put by the Principle of Relativity enforces the Galilean transformation between IRFs likewise in SR -- adding the constraint about the invariance of the speed of light in vacuum-- it enforces the Lorentz transformation.

Does it make sense ?

 

[atyy]

It depends on what one means by preserving the form of the laws. By passive transformation, one often means coordinate transformations. In Newtonian physics, coordinate transformations from inertial frames to noninertial coordinate systems also preserve the form of the laws - if one allows the laws to contain Christoffel symbols.

So if you allow Christoffel symbols, then arbitrary coordinate transformations preserve the form of the laws, and not only transformations between inertial frames. In this sense, passive transformations do not contain the same information as active transformations.

If one does not allow Christoffel symbols, then arbitarary coordinate transformations do not preserve the form of the laws, and only transformations between inertial frames do. In this sense, the passive transformations describe the same symmetry as active transformations.

 

[stevendaryl]

It depends on what one means by preserving the form of the laws. By passive transformation, one often means coordinate transformations. In Newtonian physics, coordinate transformations from inertial frames to noninertial coordinate systems also preserve the form of the laws - if one allows the laws to contain Christoffel symbols.

I spent some time thinking about rewriting Newton’s laws in a generally covariant way. Of course Cartan had already done it a long time ago, but I like to see how far I can get on my own...

What I found, which is sort of interesting, maybe, is that in order to allow for accelerated coordinate systems, you pretty much have to take a “spacetime” view. Only if time is treated like a fourth coordinate do the “g-forces” become Christoffel symbols.

 

[stevendaryl]

My approach is to start with Newton’s laws of motion in an inertial coordinate system:

$\dfrac{d P^j}{dt} = F^j$

Now, introduce a 4th coordinate $x^0$, which I will assume is just linearly related to $t$, so letting $P^0 = m \dfrac{dx^0}{dt}$, we have

$\dfrac{d P^0}{dt} = F^0 \equiv 0$

These four equations can be combined into a single 4-d vector equation:

$\dfrac{d P^\mu}{dt} = F^\mu$

This is only valid in an inertial coordinate system. But if we do a coordinate transformation from $x^\mu$ to $x^\alpha$, we have:

$\dfrac{d P^\alpha}{dt} + \Gamma^\alpha_{\beta \gamma} U^\beta P^\gamma = F^\alpha$

Where if we let $\Lambda^\alpha_\mu = \dfrac{\partial x^\alpha}{\partial x^\mu}$ and
$\Lambda^\mu_\alpha = \dfrac{\partial x^\mu}{\partial x^\alpha}$, then

$P^\alpha \equiv \Lambda^\alpha_\mu P^\mu$

$F^\alpha \equiv \Lambda^\alpha_\mu F^\mu$

$\Gamma^\alpha_{\beta \gamma} \equiv \Lambda^\alpha_\mu \dfrac{\partial}{\partial x^\beta} \Lambda^\mu_\gamma$

and

$U^\beta \equiv \dfrac{dx^\beta}{dt}$

Unlike the case in GR, the connection coordinates here are not defined in terms of derivatives of the metric, since in Newtonian physics there is no metric relating points at different times.

This generally covariant form of Newton’s laws is equivalent to the original, so there is no physical content to it. But it’s interesting that in this form, the fictitious forces of Coriolis forces, Centrifugal forces and g forces are all captured by the Christoffel symbols.

 

[haushofer]

My approach is to start with Newton’s laws of motion in an inertial coordinate system:

$\dfrac{d P^j}{dt} = F^j$

Now, introduce a 4th coordinate $x^0$, which I will assume is just linearly related to $t$, so letting $P^0 = m \dfrac{dx^0}{dt}$, we have

$\dfrac{d P^0}{dt} = F^0 \equiv 0$

These four equations can be combined into a single 4-d vector equation:

$\dfrac{d P^\mu}{dt} = F^\mu$

This is only valid in an inertial coordinate system. But if we do a coordinate transformation from $x^\mu$ to $x^\alpha$, we have:

$\dfrac{d P^\alpha}{dt} + \Gamma^\alpha_{\beta \gamma} U^\beta P^\gamma = F^\alpha$

Where if we let $\Lambda^\alpha_\mu = \dfrac{\partial x^\alpha}{\partial x^\mu}$ and
$\Lambda^\mu_\alpha = \dfrac{\partial x^\mu}{\partial x^\alpha}$, then

$P^\alpha \equiv \Lambda^\alpha_\mu P^\mu$

$F^\alpha \equiv \Lambda^\alpha_\mu F^\mu$

$\Gamma^\alpha_{\beta \gamma} \equiv \Lambda^\alpha_\mu \dfrac{\partial}{\partial x^\beta} \Lambda^\mu_\gamma$

and

$U^\beta \equiv \dfrac{dx^\beta}{dt}$

Unlike the case in GR, the connection coordinates here are not defined in terms of derivatives of the metric, since in Newtonian physics there is no metric relating points at different times.

This generally covariant form of Newton’s laws is equivalent to the original, so there is no physical content to it. But it’s interesting that in this form, the fictitious forces of Coriolis forces, Centrifugal forces and g forces are all captured by the Christoffel symbols.

Yes, and it's because these forces are linear in m. Just like Newtonian gravity ;)

I think this also bothered Einstein: how unique are general-covariant formulations? See also my notes which contain the well-known general-covariantization of the diffusion eqn.

 

[haushofer]

Thus mathematically we can state the Principle of Relativity (both in Newton mechanics and SR) in that two equivalent forms.
Then in Newton mechanics, starting from $$F = m \frac {dx^2} {dt^2} $$ supposed to be valid in a IRF, the requirement put by the Principle of Relativity enforces the Galilean transformation between IRFs likewise in SR -- adding the constraint about the invariance of the speed of light in vacuum-- it enforces the Lorentz transformation.

Does it make sense ?

Yes, but I would add a subtlety. In Newtonian spacetime the speed of light is from a certain point of view also the same for all IFR's. It's just infinitely large.

 

[PeterDonis]

In Newtonian spacetime the speed of light is from a certain point of view also the same for all IFR's. It's just infinitely large.

No, the speed of light is still finite in Newtonian spacetime--and frame-dependent.

One could say that there is an infinitely large "invariant speed" in Newtonian spacetime, but light does not travel at that speed. (Gravity, to the extent it can be said to "travel" in Newtonian physics, does travel at that speed, however, since it is an instantaneous action at a distance.) It was already known in Newton's time that light traveled with a finite speed.

 

[haushofer]

No, the speed of light is still finite in Newtonian spacetime--and frame-dependent.

One could say that there is an infinitely large "invariant speed" in Newtonian spacetime, but light does not travel at that speed. (Gravity, to the extent it can be said to "travel" in Newtonian physics, does travel at that speed, however, since it is an instantaneous action at a distance.) It was already known in Newton's time that light traveled with a finite speed.

Well, yes,that's why I say "from a certain perspective", namely the causal structure of spacetime. That's what we're discussing here. The subtlety is the double rôle of light: being a dynamical field, and determining the causal structure of spacetime.

In the correspondence principle Newtonian spacetime and its symmetries follow, loosely speaking, from a c-->oo limit/contraction.

I'd say it's problematic from a dynamical point of view to talk about a "finite speed of light in Newtonian spacetime; i don't know of any galilean-covariant formulation of Maxwell's equations which describe light propagating with a finite speed, but that's another subtlety for another topic.

 

[PeterDonis]

The subtlety is the double rôle of light: being a dynamical field, and determining the causal structure of spacetime.

But light doesn't "determine the causal structure of spacetime", in either Newtonian physics or relativity. In relativity, the behavior of light, since it happens to be massless, demonstrates the causal structure of spacetime; but it doesn't determine it. And in Newtonian physics, light has nothing to do with the causal structure of spacetime at all.

In the correspondence principle Newtonian spacetime and its symmetries follow, loosely speaking, from a c-->oo limit/contraction.

Only in some respects. For example, the Newtonian description of light is not Maxwell's Equations in the $c \rightarrow \infty$ limit. See further comments below.

i don't know of any galilean-covariant formulation of Maxwell's equations which describe light propagating with a finite speed

There isn't one, but so what? Newtonian physics did not use Maxwell's Equations to describe light. Yes, we know now that Maxwell's Equations do describe light, and that they are Lorentz invariant, not Galilean invariant; but Newtonian physics did not know that, and it did not have a fundamental theoretical description of light, only a phenomenological one. But our current physics also has only phenomenological descriptions of many things. (Someone who believes that there is some different underlying fundamental theory that gives rise to both our current GR and our current QM as approximations might say that all of our current theories of physics are only "phenomenological".) The problem with the Newtonian physics description of light was not that it was only "phenomenological", but that it was wrong--it made incorrect predictions. But it was what it was, incorrect predictions and all, as was Newtonian physics in general.

 

[vanhees71]

Unlike the case in GR, the connection coordinates here are not defined in terms of derivatives of the metric, since in Newtonian physics there is no metric relating points at different times.

This generally covariant form of Newton’s laws is equivalent to the original, so there is no physical content to it. But it’s interesting that in this form, the fictitious forces of Coriolis forces, Centrifugal forces and g forces are all captured by the Christoffel symbols.

The analogue of this Newtonian derivation of a generally covariant formalism is more like the $n$-bein formalism to "derive" GR from SR by making Poincare transformations local.

 

[haushofer]

But light doesn't "determine the causal structure of spacetime", in either Newtonian physics or relativity. In relativity, the behavior of light, since it happens to be massless, demonstrates the causal structure of spacetime; but it doesn't determine it. And in Newtonian physics, light has nothing to do with the causal structure of spacetime at all.

What I mean by that, is that the value of c determines the light cones. If you take the c --> oo limit, the future and past light cones stretch out to become all of spacetime, which is an indication of the absolute nature of time in Newtonian physics.

Only in some respects. For example, the Newtonian description of light is not Maxwell's Equations in the $c \rightarrow \infty$ limit. See further comments below.

Of course, Newton didn't look at it that way. But with "Newtonian" I mean "In the Newtonian limit". You certainly can write down the non-relativistic versions of the Maxwell equations, and these equations can be considered to be a simple c --> oo limit of the relativistic ones.

(Someone who believes that there is some different underlying fundamental theory that gives rise to both our current GR and our current QM as approximations might say that all of our current theories of physics are only "phenomenological".)

I consider these kinds of phenomena from an effective/"Wilsonian" point of view at the level of the equations of motion; different theories are then connected by the correspondence principle. Maybe my usage of "Newtonian" is confusing, and maybe I should replace it with "nonrelativistic limits of".

 

[haushofer]

Unlike the case in GR, the connection coordinates here are not defined in terms of derivatives of the metric, since in Newtonian physics there is no metric relating points at different times.

You can introduce a metrics-compatible connection. Note the plural: if you introduce a temporal and spatial metric, and their inverses via projective relations, metric compatibility determines the connection up to a closed 2-form.

These temporal and spatial metrics are covariantly-constant; without gravity, these are the two metrics which are kept invariant under Galilei-transformations.

 

[PeterDonis]

What I mean by that, is that the value of c determines the light cones.

If by "c" you mean "the invariant speed", yes. But there is no logical necessity for light to travel at the invariant speed (and in Newtonian mechanics, it doesn't).

 

[robphy]

You can introduce a metrics-compatible connection. Note the plural: if you introduce a temporal and spatial metric, and their inverses via projective relations, metric compatibility determines the connection up to a closed 2-form.

These temporal and spatial metrics are covariantly-constant; without gravity, these are the two metrics which are kept invariant under Galilei-transformations.

Are you using Ehlers’ frame theory?
Or some other formalism?

 

[haushofer]

Are you using Ehlers’ frame theory?
Or some other formalism?

To be honest, I'm not into the nomenclature; I'd just call it Newton-Cartan theory ;) But Ehlers played a role in its development, so you could be right.

See also our Insight about Newton Cartan theory here on PF and references therein.

 

[haushofer]

My approach is to start with Newton’s laws of motion in an inertial coordinate system:

...

Maybe it's nice to say something about how to derive the geodesic equation in Newton-Cartan theory, because this stuff is not well-known.

In Newton-Cartan theory you work with a spatial and a temporal metric, which are both degenerate. One has the following projective relation,

$$h^{\mu\nu}h_{\nu\rho} + \tau^{\mu\nu}\tau_{\nu\rho} = \delta^{\mu}_{\rho}$$

where h is the spatial metric (with rank 3) and tau the temporal one (with rank 1). Having rank one, we can also write

$$\tau_{\mu\nu} = \tau_{\mu}\tau_{\nu}$$

Now, the Lagrangian for a relativistic particle is well-known. The non-relativistic particle however is described by a Lagrangian which, naively, takes the following form (putting the mass to 1):

$$L = \frac{h_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}{\tau_{\rho}\dot{x}^{\rho}}$$

If you vary the corresponding action w.r.t. x, one indeed gets a geodesic equation, with a Christoffel-like connection for both metrics h and tau. But there is something missing. Relativistically, the point particle Lagrangian is not only invariant w.r.t. gct's, but also under Local Lorentz transformations (LLT's). But if you look at the Galilei (or Bargmann) algebra, you can deduce that the metric h is not invariant with respect to local Galilei boosts. The way to fix this, is to add an extra coupling:

$$L = \frac{h_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}{\tau_{\rho}\dot{x}^{\rho}} - 2m_{\mu}\dot{x}^{\mu}$$

This vector field $m_{\mu}$ is the gauge field of the central extension of the Galilei algebra, and according to the algebra transforms under local boosts. One can then show that this Lagrangian is invariant under local Galilei boosts,

$$\delta_B L = 0$$

and that the variation of the corresponding action gives one the geodesic equation of Newton-Cartan theory, where the connection now depends on h, tau and m. Ultimately, after a whole lot of gauge fixing, the Newton potential arises from the vector field $m_{\mu}$. It's kind of amusing to see that the coupling of a particle to Newtonian gravity in the Newton Cartan formalism is similar to the coupling of a relativistic particle to the electromagnetic field. This is how the Newton potential for gravity is the gauge field belonging to the generator of the central extension ("mass") of the Galilei algebra. See also page 75 and further from

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiRn9qGwJ7vAhUDrqQKHb2cAHkQFjABegQIARAD&url=https://www.rug.nl/research/portal/files/34926446/Complete_thesis.pdf&usg=AOvVaw2GAhJAD69j8f-S0aGy2o1c

 

[cianfa72]

By the way...in the context of invariant form of physical laws, I'm struggling with the tensor notation for matrices.

Take for instance the Lorentz transformation for spacetime flat metric:
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}$$then defining the following matrices:$${\Lambda}=({\Lambda^{\mu}}_{\rho}), {\eta}=(\eta_{\rho \sigma})$$the previous formula translates to: $${\Lambda}^{\text{T}} {\eta} {\Lambda}={\eta}$$
Could you help me in understanding the steps to get it ? Thanks

 

[cianfa72]

By the way...in the context of invariant form of physical laws, I'm struggling with the tensor notation for matrices.

Take for instance the Lorentz transformation for spacetime flat metric:
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}$$then defining the following matrices:$${\Lambda}=({\Lambda^{\mu}}_{\rho}), {\eta}=(\eta_{\rho \sigma})$$the previous formula translates to: $${\Lambda}^{\text{T}} {\eta} {\Lambda}={\eta}$$
Could you help me in understanding the steps to get it ? Thanks

Any help ? Thanks

 

[vanhees71]

It's just how the matrix multiplication is defined. The main problem is that you loose the nice "mnemonics" about the transformation properties from the vertical index position in the Ricci calculus. With your definition of the matrices as in your 2nd line it should be clear that
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} = (\Lambda^{\text{T}} \eta)_{\rho \nu} ,$$
and from that
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho}{\Lambda^{\nu}}_{\sigma} = (\Lambda^{\text{T}} \eta \Lambda)_{\rho \sigma}=\eta_{\rho \sigma},$$
and now you can just cancel the indices and write it just in the matrix notation:
$$\Lambda^{\text{T}} \eta \Lambda=\eta.$$

 

[cianfa72]

It's just how the matrix multiplication is defined. The main problem is that you loose the nice "mnemonics" about the transformation properties from the vertical index position in the Ricci calculus. With your definition of the matrices as in your 2nd line it should be clear that
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} = (\Lambda^{\text{T}} \eta)_{\rho \nu} ,$$

I believe that should be also equal to:
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} = (\eta^{\text{T}} \Lambda)_{\nu \rho} $$
and from that we find again
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho}{\Lambda^{\nu}}_{\sigma} = (\Lambda^{\text{T}} \eta \Lambda)_{\rho \sigma}=\eta_{\sigma \rho}$$

As far as I can understand basically we are free to assign the (upper or lower) vertical position for row and column indices (the first and the second index respectively) when representing a matrix in index notation. The other way around you have to stick to the (upper or lower) "vertical" position of the tensor index when representing it using matrices.

Any help is really appreciated.

 

[stevendaryl]

In Newton-Cartan theory you work with a spatial and a temporal metric, which are both degenerate. One has the following projective relation,

$$h^{\mu\nu}h_{\nu\rho} + \tau^{\mu\nu}\tau_{\nu\rho} = \delta^{\mu}_{\rho}$$

where h is the spatial metric (with rank 3) and tau the temporal one (with rank 1). Having rank one, we can also write

$$\tau_{\mu\nu} = \tau_{\mu}\tau_{\nu}$$

I don't understand how the spatial metric is a tensor, since the spatial distance between two spacetime points is undefined unless the temporal distance is zero. What is the meaning of $h_{\mu \nu} x^\mu x^\nu$ when the 4-vector $x^\nu$ has a nonzero time component?

 

[vanhees71]

I believe that should be also equal to:
ημνΛμρ=(ηTΛ)νρ
and from that we find again
ημνΛμρΛνσ=(ΛTηΛ)ρσ=ησρ

As far as I can understand basically we are free to assign the (upper or lower) vertical position for row and column indices (the first and the second index respectively) when representing a matrix in index notation. The other way around you have to stick to the (upper or lower) "vertical" position of the tensor index when representing it using matrices.

Any help is really appreciated.

Sure you can also start from this relation, and you always get the same formula. I don't understand, what problem you have.

 

[haushofer]

I don't understand how the spatial metric is a tensor, since the spatial distance between two spacetime points is undefined unless the temporal distance is zero. What is the meaning of $h_{\mu \nu} x^\mu x^\nu$ when the 4-vector $x^\nu$ has a nonzero time component?

It doesn't have a meaning of an invariant spacetime length, because there is no such thing nonrelativistiscally. So I can't give you a nice geometric interpretation of it based on relativistic intuition.

The starting point in Newton-Cartan theory (in the metrical language) are two (flat) metrics: a spatial and a temporal one. One can show that the two metrics which are kept invariant under the Galilei group, are

$$h^{\mu\nu} \,, \ \ \ h^{00} = h^{0j} = 0, \ \ \ h^{ij} = \delta^{ij}$$

and

$$\tau_{\mu\nu} \,, \ \ \ \tau_{0j} = \tau_{ij} = , \ \ \tau_{00} = 1$$

The "inverse metrics" are defined projectively via

$$h^{\mu\nu}h_{\nu\rho} + \tau^{\mu\nu}\tau_{\nu\rho} = \delta^{\mu}_{\rho}$$

In Newton-Cartan theory these metrics become dynamical. The metric compatibility is imposed on our two metrics $h^{\mu\nu}$ and $\tau_{\mu\nu}$, which define a connection up to a closed 2-form. The other two metrics $h_{\mu\nu}$ and $\tau^{\mu\nu}$ are (as one can check) NOT metric-compatibel (i.e. the covariant derivative of these metrics doesn't vanish).

The (non)tensorial character of these metrics and connection is not different from those in GR, because the defining relations are tensor equations under gct's.

Hope this helps. Maybe it's time to write a more detailed Insight about Newton-Cartan theory :P

 

[cianfa72]

Sure you can also start from this relation, and you always get the same formula. I don't understand, what problem you have.

I was wondering if there is let's say a "rule" to "recover" the 'vertical' index position (upper or lower) when transforming a matrix in the corresponding tensor.
For instance...starting from the matrix product $(\Lambda^{\text{T}} \eta)$ representing the tensor $\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho}$ based on which 'rule' we'll choose to transform it back in the form
$$(\Lambda^{\text{T}} \eta)_{\rho \nu}$$ rather than into ${(\Lambda^{\text{T}} \eta)_{\rho}}^{\nu}$ ?

 

[vanhees71]

That's the problem with the matrix notation. You have no good way to keep track of whether a scheme of numbers are co- or contravariant tensor components. If in doubt, I think the Ricci calculus with clearly vertically and horizontally placed indices is a more save way to calculate.

 

[cianfa72]

If in doubt, I think the Ricci calculus with clearly vertically and horizontally placed indices is a more save way to calculate.

That includes also staggered indices, I believe. Therefore the following
$$(\Lambda^{\text{T}} \eta)^{\rho}_{\nu}$$
is not formally correct.

 

[vanhees71]

As I said, you must take care of both the "vertical and horizontal placement" of the indices (an exception are symmetric tensors of course, where the horizontal placement doesn't matter).