# I Why is energy not Lorentz invariant?

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1. Jul 30, 2016

### Frank Castle

As I understand it, since space-time is modelled 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)?

2. Jul 30, 2016

### robphy

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

3. Jul 30, 2016

### Frank Castle

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)?!

4. Jul 30, 2016

### Ibix

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

5. Jul 30, 2016

### Frank Castle

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.

6. Jul 30, 2016

### 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)

7. Jul 30, 2016

### Ibix

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.

8. Jul 30, 2016

### Frank Castle

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 $p^a u_{Alice}^b g_{ab}$?

Last edited: Jul 30, 2016
9. Jul 30, 2016

### robphy

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

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).

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").

10. Jul 30, 2016

### DrGreg

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.

11. Jul 30, 2016

### Staff: Mentor

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.

12. Jul 31, 2016

### vanhees71

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.

13. Jul 31, 2016

### PeroK

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.

14. Jul 31, 2016

### Frank Castle

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

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).

Last edited: Jul 31, 2016
15. Jul 31, 2016

### 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.

Last edited: Jul 31, 2016
16. Jul 31, 2016

### Frank Castle

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?!

17. Jul 31, 2016

### 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.

18. Aug 1, 2016

### Markus Hanke

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.

19. Aug 1, 2016

### vanhees71

For a more conventional treatment of special-relativistic mechanics see

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

$$L_0=-m c^2 \sqrt{1-\vec{v}^2/c^2},$$
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.

20. Aug 1, 2016

### Frank Castle

How does one derive them without using 4-vectors?

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).