I Why is energy not Lorentz invariant?

[Frank Castle]

As I understand it, since space-time is modeled as a four dimensional manifold it is natural to consider 4 vectors to describe physical quantities that have a direction associated with them, since we require that physics should be independent of inertial frame and so we should describe it in terms of objects that are invariant under Lorentz transformations.

Having said this, I'm slightly unsure as to why certain quantities that are scalars in classical mechanics are no longer scalars in the context of special relativity? For example, what is the reason why energy is not Lorentz invariant (I get that it is combined with 3 momentum to construct a 4 momentum vector, but I don't really understand why)?

 

[robphy]

Kinetic energy in PHY 101 isn't Galilean invariant... but it is rotationally invariant.

 

[Frank Castle]

Kinetic energy in PHY 101 isn't Galilean invariant... but it is rotationally invariant.

Yes, that's true, but nonetheless energy is a scalar in classical mechanics (CM) whereas in the framework of special relativity (SR) it is not. Is the reason for this purely because in CM a scalar is defined as a quantity that is invariant under spatial rotations and reflections, whereas in SR a scalar is defined as a quantity that is invariant under Poincare transformations, which includes Lorentz boosts, which are essentially the generalisation of Galilean boosts (which, as you say, energy is already not an invariant quantity under even in CM)?!

 

[Ibix]

Energy is still a scalar - it's only one component of a 4-vector, not a 4-vector itself.

 

[Frank Castle]

Energy is still a scalar - it's only one component of a 4-vector, not a 4-vector itself.

But it is now a component of a 4-vector and so it transforms between frames $$p'^{0}=\Lambda^{0}_{\;\mu}p^{\mu}\Rightarrow E'=\Lambda^{0}_{\;0}E +\Lambda^{0}_{\;i}p^{i}$$ so it isn't a Lorentz scalar.

 

[robphy]

One has to distinguish
"observer-dependent scalar" (like $p^a u_{Alice}^b g_{ab}$, which depends on Alice's 4-velocity $u$)
from observer-independent scalars (like $p^a p^bg_{ab}$).

Everyone will agree that Alice measures $p^a u_{Alice}^b g_{ab}$...
but everyone agrees that Bob will measure something different $p^a u_{Bob}^b g_{ab}$ .

Everyone will agree that everyone measures the same value for $p^a p^bg_{ab}$.

(sorry numerous edits...
meant: observer-dependent not observer-invariant... clarified observer-independent)

 

[Ibix]

But it is now a component of a 4-vector and so it transforms between frames $$p'^{0}=\Lambda^{0}_{\;\mu}p^{\mu}\Rightarrow E'=\Lambda^{0}_{\;0}E +\Lambda^{0}_{\;i}p^{i}$$ so it isn't a Lorentz scalar.

Indeed. But, as robphy pointed out, it transforms between frames in Newtonian physics too, so that fact doesn't seem particularly surprising to me.

Edit: that sounds snarkier than I intended - sorry. I don't quite get why you are surprised, however.

 

[Frank Castle]

One has to distinguish
"observer-dependent scalar" (like $p^a u_{Alice}^b g_{ab}$, which depends on Alice's 4-velocity $u$)
from observer-independent scalars (like $p^a p^bg_{ab}$).

Everyone will agree that Alice measures $p^a u_{Alice}^b g_{ab}$...
but everyone agrees that Bob will measure something different $p^a u_{Bob}^b g_{ab}$ .

Everyone will agree that everyone measures the same value for $p^a p^bg_{ab}$.

(sorry numerous edits...
meant: observer-dependent not observer-invariant... clarified observer-independent)

Having thought about it, like you said, energy is observer dependent even in classical mechanics. Is the point that scalar quantities do not depend on the coordinates one chooses in one's frame of reference, however, they do in general depend on one's frame of reference?! There are, however, scalar quantities which are also independent of one's reference frame, such as an objects mass, or the distance between two objects.
I think what has confused me is that, at least from what I've read, this isn't really emphasised in textbooks. In what I've read on special relativity, there is a lot of emphasis put on objects that are Lorentz scalars, but not on the fact that one can have observer dependent scalars.

In the formula that you've written, is $p^{a}$ the 4-momentum of an object that Alice/Bob is observing (and as such, the scalar is frame dependent because, in general, Alice and Bob will have different 4-velocities)?

By the way, how does one drive the formula $p^a u_{Alice}^b g_{ab}$?

 

[robphy]

So energy is observer dependent even in classical mechanics then? I think what has confused me is that, at least from what I've read, this isn't really emphasised in textbooks. In what I've read on special relativity, there is a lot of emphasis put on objects that are Lorentz scalars, but not on the fact that one can have observer dependent scalars.

I saw there was some confusion brewing... so I wrote down some explicit expressions to clarify.

In the formula that you've written, is $p^{a}$ the 4-momentum of an object that Alice/Bob is observing (and as such, the scalar is frame dependent because, in general, Alice and Bob will have different 4-velocities)?

Yes... I'd say for extreme clarity that
everyone agrees on what Alice measures for the energy of the particle and
everyone agrees on what Bob measures for the energy of the particle,
and everyone agrees that Alice's and Bob's measured values for the energy are different (in general).

By the way, how does one drive the formula $p^a u_{Alice}^b g_{ab}$?

This is fancy way of saying "take the dot-product of the particle 4-momentum and the measuring observer's [unit] 4-velocity (i.e. "take the time-component of the particle 4-momentum").

 

[DrGreg]

By the way, how does one drive the formula $p^a u_{Alice}^b g_{ab}$?

What would the value be in Alice's own local coordinates? As the expression is a valid invariant scalar, it takes that value in all coordinate systems.

 

[Nugatory]

Is the point that scalar quantities do not depend on the coordinates one chooses in one's frame of reference, however, they do in general depend on one's frame of reference?! There are, however, scalar quantities which are also independent of one's reference frame, such as an objects mass, or the distance between two objects.

The word "scalar" is used in two different senses, and unfortunately you're generally expected to figure out which one it is from the context.

One meaning is "not a vector". These are not necessarily either Galilean or Lorentz-invariant (although they might be). Kinetic energy is an example - it is egregiously frame-dependent.

The the other meaning is "rank-zero tensor", and these are necessarily coordinate-independent invariants.

All scalars by the second definition are scalars by the first, but not the other way around.

 

[vanhees71]

Yes, that's true, but nonetheless energy is a scalar in classical mechanics (CM) whereas in the framework of special relativity (SR) it is not. Is the reason for this purely because in CM a scalar is defined as a quantity that is invariant under spatial rotations and reflections, whereas in SR a scalar is defined as a quantity that is invariant under Poincare transformations, which includes Lorentz boosts, which are essentially the generalisation of Galilean boosts (which, as you say, energy is already not an invariant quantity under even in CM)?!

Kinetic energy is a scalar under rotations in both Newtonian mechanics and special relativity. It is neither a scalar under Galileo nor Lorentz boosts. In physics you have always consider the group of transformations with respect to which you define the covariance of the corresponding objects.

 

[PeroK]

Yes, that's true, but nonetheless energy is a scalar in classical mechanics (CM) whereas in the framework of special relativity (SR) it is not. Is the reason for this purely because in CM a scalar is defined as a quantity that is invariant under spatial rotations and reflections, whereas in SR a scalar is defined as a quantity that is invariant under Poincare transformations, which includes Lorentz boosts, which are essentially the generalisation of Galilean boosts (which, as you say, energy is already not an invariant quantity under even in CM)?!

You're looking at this all wrong! In classical physics, momentum and kinetic energy are clearly related (but the connection is rather vague). In SR, they are unified into the energy-momentum 4-vector. And all is beautifully clear! This should be a lightbulb moment for you.

But, instead, it's a source of confusion! I can't understand this. SR is the real deal when it comes to particle motion. You should appeciate it for what it is.

 

[Frank Castle]

The word "scalar" is used in two different senses, and unfortunately you're generally expected to figure out which one it is from the context.

One meaning is "not a vector". These are not necessarily either Galilean or Lorentz-invariant (although they might be). Kinetic energy is an example - it is egregiously frame-dependent.

The the other meaning is "rank-zero tensor", and these are necessarily coordinate-independent invariants.

All scalars by the second definition are scalars by the first, but not the other way around.

Ah ok, I guess I hadn't really appreciated this before. Thanks for highlighting this point.

You're looking at this all wrong! In classical physics, momentum and kinetic energy are clearly related (but the connection is rather vague). In SR, they are unified into the energy-momentum 4-vector. And all is beautifully clear! This should be a lightbulb moment for you.

But, instead, it's a source of confusion! I can't understand this. SR is the real deal when it comes to particle motion. You should appeciate it for what it is.

You're right, I don't really understand myself why it has been a source of confusion for me. I think it might be because I hadn't really appreciated the point that one can have quantities that are scalars, but still frame dependent, and one can also have quantities that are scalars, but are frame independent. Is the point that physical quantities can have no directional dependence, but this doesn't mean that they are invariant when one transforms between two reference frames that are in relative motion with respect to one another - they are still rotationally invariant, but the numerical value describing the physical quantity is different in the two frames?!

By the way, is the reason why one considers 4-vectors (as opposed to 3-vectors) in special relativity because a) the requirement of the (inertial) frame independence of the speed of light implies the Lorentz transformations between inertial frames of reference, and furthermore that space and time should be unified into a single 4-dimensional continuum (since there is no well-defined observer independent way to distinguish the two), and b) since inertial frames are related by Lorentz (or, more generally, Poincaré) transformations we require physical quantities must be covariant under Lorentz transformations, in order for physics to be (inertial) observer independent. Consequently, the fact that space-time is 4-dimensional already implies that one should consider 4-vectors and furthermore, 4-vectors transform correctly (i.e. covariantly) under Lorentz transformations (whereas 3 vectors do not - they become frame dependent quantities, and this is not satisfactory, since vectors should exist independently of reference frames).

Also, what is the justification for why energy is the zeroth component of 4-momentum? Does it simply follow from noting that $p^{\mu}=mu^{\mu}$ (where $u^{\mu}=\frac{dx^{\mu}}{d\tau}$ is the 4-velocity, with $\tau$ the proper time), and since $$u^{\mu}=\frac{dx^{\mu}}{d\tau}=\gamma\frac{dx^{\mu}}{dt}$$ where $t$ is the coordinate time in a given inertial frame, and $\gamma=\frac{dt}{d\tau}=\frac{1}{\sqrt{1-v^{2}/c^{2}}}$ is the Lorentz factor. It follows that $$p^{\mu}=m\gamma\frac{dx^{\mu}}{dt} =(mc\gamma, m\gamma\mathbf{v})$$ If one then takes the small speed, i.e. $v<<c$ limit of the zeroth component, one finds that $$p^{0}\simeq\frac{1}{c}\left(mc^{2}+\frac{1}{2}mv^{2}+\mathcal{O}(v^{2}/c^{2})\right)$$ and so we recognise the kinetic energy term suggesting that in this limit $p^{0}\simeq E/c$, and this further suggests that, in general $p^{0}=E/c$ (since, by the correspondence principle, in the non-relativistic limit, the special relativistic form should agree with the Newtonian form).

 

[Geometry_dude]

Well, there are many concepts of energy, so it is helpful to be a bit more precise what exactly one is talking about. All of them are somehow related, but there is a subtle difference.

From a dynamical point of view, the most important one is the Hamiltonian (defined on the cotangent bundle aka "momentum phase space"), which is a scalar field:
$$ H \left( q, p \right) = - \frac{m}{2} g^{ij}\, p_i \, p_j + V \left( q \right) \quad . $$
This works on general spacetimes and is always a constant of motion. Note that the $-$ sign is not accidental as this is the $(+,-,-,-)$ convention and $i,j \in \lbrace 0,1,2,3 \rbrace$. You can also take minus of that, but then you have to switch the signs in the Hamilton equations.

This in turn is just obtained from a reformulation of Newton's law on general spacetimes:
$$ F = m \nabla_v v = \text{grad} V \quad ,$$
where $\nabla$ denotes the covariant derivative and $v$ may denote a vector field or the tangent vector of a curve (smooth, timelike, future oriented, normalized to $c$).

This is different from the 'energy', that is the variable 'conjugate to time' in Minkowski spacetime:
$$E :=c p^0 = mc \dot x^0 = mc^2 \dot t $$

The latter is only a constant of motion, if there is no acceleration and - as a component of a tangent vector - not a scalar. The justification for calling it $E$ is usually given by the formula you gave. Now here comes the tricky part (as implicitly stated by Robphy): One can 'turn' this into a scalar (field over some observer curve), if one considers the vector field $(c, 0,0,0)$ on Minkowski spacetime and then takes the Minkowski product with the tangent vector of an observer (smooth, timelike, future oriented, normalized to $c$). Up to factors in $m$ and $c$, you can check that this gives you $E$ as well. But this depends strongly on the vector field I gave you. (That these are indeed 'Alice's and Bob's vector fields', if the coordinates are adapted to their respective inertial motion requires some serious justification...).

However, if you look at the Newtonian limit (which implicitly depends on the observer that is 'observing') you want to have $\dot t \approx 1$, which results in $E \approx m c^2$. So the justification usually given is not very good. If you consider H as a function on velocity phase space (for those that understand: pullback with g to the tangent bundle), you get in the Newtonian limit on Minkowski spacetime:

$$H (x, v) \approx - m c^2 + \frac{m}{2} {\vec v}^2 + V \left( x \right)$$

The additive constant makes no dynamical difference. (why there is no $t$ in the $V$ is again a bit subtle... )

I suggest learning first some Lorentzian geometry together with GR. SR makes a lot more sense then and you tend not to believe some of the nonsense one finds in textbooks.

 

[Frank Castle]

Well, there are many concepts of energy, so it is helpful to be a bit more precise what exactly one is talking about. All of them are somehow related, but there is a subtle difference.

From a dynamical point of view, the most important one is the Hamiltonian (defined on the cotangent bundle aka "momentum phase space"), which is a scalar field:
$$ H \left( q, p \right) = - \frac{m}{2} g^{ij}\, p_i \, p_j + V \left( q \right) \quad . $$
This works on general spacetimes and is always a constant of motion. Note that the $-$ sign is not accidental as this is the $(+,-,-,-)$ convention and $i,j \in \lbrace 0,1,2,3 \rbrace$. You can also take minus of that, but then you have to switch the signs in the Hamilton equations.

This in turn is just obtained from a reformulation of Newton's law on general spacetimes:
$$ F = m \nabla_v v = \text{grad} V \quad ,$$
where $\nabla$ denotes the covariant derivative and $v$ may denote a vector field or the tangent vector of a curve (smooth, timelike, future oriented, normalized to $c$).

This is different from the 'energy', that is the variable 'conjugate to time' in Minkowski spacetime:
$$E :=c p^0 = mc \dot x^0 = mc^2 \dot t $$

The latter is only a constant of motion, if there is no acceleration and - as a component of a tangent vector - not a scalar. The justification for calling it $E$ is usually given by the formula you gave. Now here comes the tricky part (as implicitly stated by Robphy): One can 'turn' this into a scalar (field over some observer curve), if one considers the vector field $(c, 0,0,0)$ on Minkowski spacetime and then takes the Minkowski product with the tangent vector of an observer (smooth, timelike, future oriented, normalized to $c$). Up to factors in $m$ and $c$, you can check that this gives you $E$ as well. But this depends strongly on the vector field I gave you. (That these are indeed 'Alice's and Bob's vector fields', if the coordinates are adapted to their respective inertial motion requires some serious justification...).

However, if you look at the Newtonian limit (which implicitly depends on the observer that is 'observing') you want to have $\dot t \approx 1$, which results in $E \approx m c^2$. So the justification usually given is not very good. If you consider H as a function on velocity phase space (for those that understand: pullback with g to the tangent bundle), you get in the Newtonian limit on Minkowski spacetime:

$$H (x, v) \approx - m c^2 + \frac{m}{2} {\vec v}^2 + V \left( x \right)$$

The additive constant makes no dynamical difference. (why there is no $t$ in the $V$ is again a bit subtle... )

I suggest learning first some Lorentzian geometry together with GR. SR makes a lot more sense then and you tend not to believe some of the nonsense one finds in textbooks.

Thanks for your detailed answer. I've found from reading introductory texts on special relativity, this distinction isn't really discussed. Would there be any texts that you would recommend in particular?

Also, would what I put in the latter half of my previous post (#14) be correct reasoning all for motivating why we consider space and time as a 4-dimensional space-time, and why we use 4-vectors?!

 

[SiennaTheGr8]

Frank Castle: You can derive the relativistic equations for energy and momentum without using four-vectors. If you do that, then you don't need to "justify" labeling the time component of the four-momentum "energy"; it just is, because the quantity $\gamma m c^2$ appears, and you already recognize it.

Also, rest energy (mass) is indeed a Lorentz-invariant scalar.

 

[Markus Hanke]

For example, what is the reason why energy is not Lorentz invariant

Perhaps it would help if you turn things around a bit, and ask yourself - what are the fundamental conserved quantities in a region of Minkowski spacetime ? The answer is given by Noether's theorem - to every differentiable symmetry generated by a local action, there corresponds a conserved current. So far as energy is concerned, the relevant symmetry is time-translation invariance; but since time is an observer-dependent concept in SR, then so is the resulting conserved quantity. That is why energy is observer-dependent.

To obtain something that is truly covariant, we need to widen our perspective a little, and treat both time and space on equal footing - i.e. we don't just look at time-translation invariance as a symmetry, but rather at translations in spacetime instead. The conserved quantity associated with the Noether current of this symmetry is the full energy-momentum tensor. Energy density is one of the components of this tensor, but once again, individual tensor components are not Lorentz invariant, only the overall tensor itself is.

Note that this way of looking at things does not make explicit reference to how you label the individual components of the various objects; it looks only at the relationships between them.

 

[vanhees71]

Thanks for your detailed answer. I've found from reading introductory texts on special relativity, this distinction isn't really discussed. Would there be any texts that you would recommend in particular?

Also, would what I put in the latter half of my previous post (#14) be correct reasoning all for motivating why we consider space and time as a 4-dimensional space-time, and why we use 4-vectors?!

For a more conventional treatment of special-relativistic mechanics see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The usual free Lagrangian reads
$$L_0=-m c^2 \sqrt{1-\vec{v}^2/c^2},$$
and the corresponding Hamiltonian
$$H_0=\vec{p} \cdot \vec{v}-L_0=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}=c \sqrt{m^2 c^2+\vec{p}^2},$$
where
$$\vec{p}=\frac{\partial L_0}{\partial v}=\frac{ m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}$$
is the momentum of the particle.

You can reformulate this also in manifestly covariant form. For that, see the above linked article.

 

[Frank Castle]

Frank Castle: You can derive the relativistic equations for energy and momentum without using four-vectors. If you do that, then you don't need to "justify" labeling the time component of the four-momentum "energy"; it just is, because the quantity $\gamma m c^2$ appears, and you already recognize it.

Also, rest energy (mass) is indeed a Lorentz-invariant scalar.

How does one derive them without using 4-vectors?

Perhaps it would help if you turn things around a bit, and ask yourself - what are the fundamental conserved quantities in a region of Minkowski spacetime ? The answer is given by Noether's theorem - to every differentiable symmetry generated by a local action, there corresponds a conserved current. So far as energy is concerned, the relevant symmetry is time-translation invariance; but since time is an observer-dependent concept in SR, then so is the resulting conserved quantity. That is why energy is observer-dependent.

That makes a lot of sense. So is the point that energy is the conserved quantity corresponding to time translation invariance, however, the full symmetry in special relativity is space-time translation invariance (quantities are not invariant under either space or time independently, but simultaneously under both).

For a more conventional treatment of special-relativistic mechanics see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The usual free Lagrangian reads
$$L_0=-m c^2 \sqrt{1-\vec{v}^2/c^2},$$
and the corresponding Hamiltonian
$$H_0=\vec{p} \cdot \vec{v}-L_0=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}=c \sqrt{m^2 c^2+\vec{p}^2},$$
where
$$\vec{p}=\frac{\partial L_0}{\partial v}=\frac{ m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}$$
is the momentum of the particle.

You can reformulate this also in manifestly covariant form. For that, see the above linked article.

Thank you for the notes, I've had a read and they've been helpful in my understanding.

Would the correct reasoning why one considers 4-vectors (as opposed to 3-vectors) in special relativity because a) the requirement of the (inertial) frame independence of the speed of light implies the Lorentz transformations between inertial frames of reference, and furthermore that space and time should be unified into a single 4-dimensional continuum (since there is no well-defined observer independent way to distinguish the two), and b) since inertial frames are related by Lorentz (or, more generally, Poincaré) transformations we require physical quantities must be covariant under Lorentz transformations, in order for physics to be (inertial) observer independent. Consequently, the fact that space-time is 4-dimensional already implies that one should consider 4-vectors and furthermore, 4-vectors transform correctly (i.e. covariantly) under Lorentz transformations (whereas 3 vectors do not - they become frame dependent quantities, and this is not satisfactory, since vectors should exist independently of reference frames).

 

[vanhees71]

I'd say, it's simply for convenience that you formulate physical laws (equations of motion of all kind) in a manifestly covariant way. Special-relativistic spacetime is the Minkowski space, and a physical laws for closed systems should be fully Poincare invariant for consistency with that space-time model. To find such laws, it's most convenient to use Minkowski tensors (and tensor fields), because they have a well defined and pretty simple (linear) transformation behavior.

An exception is the principle of least action of point particles. Here, it is simpler to use the (1+3) formalism (see my manuscript linked above). However, in this case the action in the Lagrange form is manifestly a Lorentz scalar and thus the equations of motion are also consistent with the Minkowski-space "geometry". To see this explicitly you rather define covariant quantities. For (kinetic) energy and momentum of a point particle it's obviously the time derivatives which make trouble, because time is a frame-dependent quantity, but there's a preferred frame for massive particles, namely the instantaneous rest frame of the particle, and you can define the "proper time" of the particle, by summing up the time increments in those local rest frames. For the "time derivatives" you only need the increment, and it's defined by (setting $c=1$ for simplicity)
$$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2}.$$
Dhis is obviously invariant, because you have
$$\mathrm{d} \tau^2 = \eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
Then you define
$$p^{\mu} = \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=\frac{m}{\sqrt{1-v^2}} \begin{pmatrix} 1 \\ \vec{v} \end{pmatrix}.$$

 

[Frank Castle]

I'd say, it's simply for convenience that you formulate physical laws (equations of motion of all kind) in a manifestly covariant way. Special-relativistic spacetime is the Minkowski space, and a physical laws for closed systems should be fully Poincare invariant for consistency with that space-time model. To find such laws, it's most convenient to use Minkowski tensors (and tensor fields), because they have a well defined and pretty simple (linear) transformation behavior.

An exception is the principle of least action of point particles. Here, it is simpler to use the (1+3) formalism (see my manuscript linked above). However, in this case the action in the Lagrange form is manifestly a Lorentz scalar and thus the equations of motion are also consistent with the Minkowski-space "geometry". To see this explicitly you rather define covariant quantities. For (kinetic) energy and momentum of a point particle it's obviously the time derivatives which make trouble, because time is a frame-dependent quantity, but there's a preferred frame for massive particles, namely the instantaneous rest frame of the particle, and you can define the "proper time" of the particle, by summing up the time increments in those local rest frames. For the "time derivatives" you only need the increment, and it's defined by (setting $c=1$ for simplicity)
$$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2}.$$
Dhis is obviously invariant, because you have
$$\mathrm{d} \tau^2 = \eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
Then you define
$$p^{\mu} = \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=\frac{m}{\sqrt{1-v^2}} \begin{pmatrix} 1 \\ \vec{v} \end{pmatrix}.$$

I was just wondering if what I wrote would be a correct intuitive argument for why one would use 4-vectors and why we consider spacetime, instead of space and time separately. The latter I assume is the case because in Newtonian mechanics, time is treated as absolute, i.e. a frame independent quantity, and thus one can construct a 3-dimensional space at a particular instant in time, and because absolute simultaneity exists, every observer will agree on this instant of time being the same. In this context, space is then treated as a collection of hyper surfaces parametrised by time (since time is a frame independent quantity in this framework, such a parametrisation of spatial hyper surfaces is frame independent, and well-defined, and (I think) unique, up to a shift in initial time). In this sense, "all of space at a given instant in time" is meaningful. Since, regardless of how one labels points in 3-D space, geometrical properties, such as the distance between two points, are independent of observer, i.e. They are invariant under rotations and Galilean transformations. However, in special relativity, this notion is no longer meaningful, since time becomes a frame dependent quantity and thus there is no observer independent way to separate time and space (in the same way as we could in Newtonian mechanics). Consequently, geometrical quantities such as the spatial distance between two points are no longer frame-independent, however, their 4 dimensional analogues, such as "space-time" distance, are. This suggests that they are not separate entities, but rather form a unified 4-dimensional continuum, i.e. space-time. Each observer in spacetime can then choose a their own set of coordinates to label points within this 4-D space-time, choosing their coordinate axes such that one coordinate is a temporal coordinate, and the 3 others are spatial coordinates.

 

[vanhees71]

That's a very nice description of the idea of special-relativistic spacetime!

 

[Frank Castle]

That's a very nice description of the idea of special-relativistic spacetime!

Thanks. So what I put about why can treat space and time separately in Newtonian mechanics is correct then?
Building on this description, would it be correct to motivate the introduction of 4-vectors since the realisation of space-time as a four-dimensional space already suggests that the natural objects to consider, that encode directional dependence of certain physical quantities, are vectors that "live" in 4-dimensional vector spaces, i.e. 4-vectors. Furthermore, as additional motivation, (according to Einstein's postulates) we require that the equations describing physical phenomena are observer independent, and thus we require the mathematical objects that we use to formulate such equations to transform covariantly between reference frames. This amounts to requiring such mathematical objects to transform covariantly under Lorentz transformations, such that their magnitudes are Lorentz invariant. For physical quantities which have directional dependence it is therefore natural to consider 4-vectors to accurately describe them in a coordinate independent manner in space-time.

 

[vanhees71]

Yes, in Newtonian mechanics the spacetime is a fiber bundle. You just put Euclidean 3D spaces along a time axis. So for Newtonian spacetime there's not much to win by lumping time and space into a four-dimensional continuum, while the natural mathematical structure to describe special-relativistic spacetime is indeed a pseudo-Euclidean 4D affine manifold, Minkowski space. The observables should than be compatible with this space-time structure, and thus be formulated covariantly (with Minkowski tensors/tensor fields).

 

[Frank Castle]

Yes, in Newtonian mechanics the spacetime is a fiber bundle. You just put Euclidean 3D spaces along a time axis. So for Newtonian spacetime there's not much to win by lumping time and space into a four-dimensional continuum, while the natural mathematical structure to describe special-relativistic spacetime is indeed a pseudo-Euclidean 4D affine manifold, Minkowski space. The observables should than be compatible with this space-time structure, and thus be formulated covariantly (with Minkowski tensors/tensor fields).

I have to admit, I'm not particularly familiar with the notion of fiber bundles, but I'm guessing what I wrote is a heuristic description of the more formal description you've put in terms of fiber bundles?!

 

[vanhees71]

Yes!

 

[Frank Castle]

Yes!

Ok, cool. Thanks for your help on that.

So, going back to the topic of scalars... Is the point essentially that one can have scalar quantities, in the sense that they have no directional dependence (i.e. they are rotationally invariant), however, they needn't necessarily be frame invariant quantities. As such, a physical quantity such as energy has no directional dependence with respect to any frame of reference and can therefore be described by a scalar, however, the numerical value of this scalar is a frame dependent quantity, inasmuch as two observers in two different reference frame will measure a different numerical (scalar) value for the energy associated with a particular object?!

 

[SiennaTheGr8]

Yeah, it sounds like you actually "get" all this, Frank Castle.

As you said a couple posts up, the four-vector formalism is motivated by the facts before us: if there's a universal speed limit (and if the laws of physics are frame-independent), then we get time dilation, length contraction, and the relativity of simultaneity. The old three-vectors are therefore no longer covariant (except in the classical limit).

All of special relativity can be done without four-vectors, but they're a convenient way to notate and conceive of the physics.

Now, you asked me how the equations for relativistic energy and momentum can be derived without four-vectors.

For starters, I'd direct you to Einstein's original "$E=mc^2$" paper: https://www.fourmilab.ch/etexts/einstein/E_mc2/www/.

If you read "between the lines," you'll see that he assumes the existence of rest energy $E_0$, a safe assumption because how else could a resting body remain at rest while losing energy? He then uses the relativistic Doppler shift (which he'd derived in his previous special-relativity paper using the Lorentz transformation: https://www.fourmilab.ch/etexts/einstein/specrel/www/index.html#SECTION22) to show that the relativistic equation for kinetic energy must be $E_k = E_0(\gamma - 1)$, which reduces to the approximation $E_k \approx \frac{1}{2}E_0 \frac{v^2}{c^2}$ in the classical limit, such that $m = E_0 / c^2$ (because classically $E_k = \frac{1}{2}mv^2$).

That last result was of course revolutionary, and so it is the explicit focus of the paper. But if you take a step back, the real meat of the paper (IMO) is the tacit introduction of the concept of rest energy, the formulation of kinetic energy in terms of it, and the implied equations for total relativistic energy $E$:

$E = E_0 + E_k\\ E = E_0 + E_0(\gamma - 1)\\ E = \gamma E_0 = \gamma mc^2.$

As for momentum ($\vec p = \gamma m \vec v$), see the famous thought experiment proposed by Gilbert Lewis and Richard Tolman in 1909, reproduced in many books on SR. For instance: https://books.google.com/books?id=FrgVDAAAQBAJ&pg=PA76

I think it's worth emphasizing that both Einstein's derivation of relativistic $E$ and Lewis/Tolman's derivation of relativistic $\vec p$ assume the conservation of energy and momentum (respectively).

You can then construct the four-vector formalism by assigning the (infinitesimal) timelike interval a "direction" in spacetime (the direction of a traveler's world line, naturally) and calling the resulting object the four-displacement: $d \vec R = (c \, dt, d \vec r)$. Differentiate $\vec R$ with respect to the traveler's proper time, and you have the four-velocity, which looks like this when simplified: $\vec V = (\gamma c, \gamma \vec v)$. Multiply $\vec V$ by the invariant $m$, and you've got the four-momentum: $\vec P = (\gamma m c, \gamma m \vec v)$, which you recognize as equivalent to $\vec P = (E / c, \vec p)$.

 

[SiennaTheGr8]

Ok, cool. Thanks for your help on that.

So, going back to the topic of scalars... Is the point essentially that one can have scalar quantities, in the sense that they have no directional dependence (i.e. they are rotationally invariant), however, they needn't necessarily be frame invariant quantities. As such, a physical quantity such as energy has no directional dependence with respect to any frame of reference and can therefore be described by a scalar, however, the numerical value of this scalar is a frame dependent quantity, inasmuch as two observers in two different reference frame will measure a different numerical (scalar) value for the energy associated with a particular object?!

Yes. As someone said on the previous page, there are unfortunately two different definitions of scalar in use. Basically:

1) a number (i.e., not a vector);

2) a quantity that is invariant and whose value is just a number.

According to the first definition, total energy, kinetic energy, and rest energy are all scalars. According to the second definition, rest energy is a scalar but total energy and kinetic energy aren't.

 

[Frank Castle]

Yes. As someone said on the previous page, there are unfortunately two different definitions of scalar in use. Basically:

1) a number (i.e., not a vector);

2) a quantity that is invariant and whose value is just a number.

According to the first definition, total energy, kinetic energy, and rest energy are all scalars. According to the second definition, rest energy is a scalar but total energy and kinetic energy aren't.

I guess one needs to be careful then when one stipulates the transformations in which a particular quantity is a scalar with respect to.

Ok cool, I think it's becoming clearer to me now. Thanks for your help, and thanks for the previous post (#29) detailing the relativistic energy derivation.

 

[Traruh Synred]

Apart from technical mathematical cases (tensors) you might just consider 'scalar' to be a real number as opposed to a set of numbers. E.,g., px is a 'scalar' but it makes no sense to ask how it transforms other then in the context of being one element of a 4-vector. Energy is just the same.

Some 'energies' are defined a non-invariant way. E.g.. temperature is given by the average energy of the particles making up a block of stuff. It is defined only in the rest from of the stuff. It would not make sense to say that a object heats up when you run pass it even though the energy of the particles making up your stuff would be higher, this average energy would not be useful in thermo dynamics where the average energy in the rest frame of the stuff is the relevant quantity.

If you were computing the scattering of one blick of stuff with others than it would be relevant to treat the energy contributed by to the total by the 'internal' motion of the particles to the four vector describing the motion of he block. This is, typically, negligible even for very hot objects.