# Is energy Galilean invariant

• B

## Main Question or Discussion Point

As the title says, is energy Galilean invariant?

I'm fairly sure it isn't, since if one considers the simple case of a free particle, such that its energy is ##E=\frac{p^{2}}{2m}##, then under a Galilean boost, it follows that ##E'= \frac{p'^{2}}{2m}=E+\frac{\tilde{p}^{2}}{2m}-\frac{\mathbf{p}\cdot\tilde{\mathbf{p}}}{m}##, where ##\mathbf{p}'=\mathbf{p}-\tilde{\mathbf{p}}##, with ##\tilde{\mathbf{p}}=m\tilde{\mathbf{v}}##, and ##\tilde{\mathbf{v}}## is the relative velocity between the two inertial frames.

I mean, it seems obvious, since momentum is not Galilean invariant (indeed, it transforms as ##\mathbf{p}'=\mathbf{p}-\tilde{\mathbf{p}}##), but I'm having a momentary crisis of confidence in my understanding and so wanted to check!

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Dale
Mentor
I'm fairly sure it isn't
You are correct, it is not invariant.

I'm having a momentary crisis of confidence
Confidence restored!

vanhees71
Ibix
You are correct. Consider the trivial case of a particle at rest in one frame. What is its kinetic energy in that frame, compared to in any other frame?

Energy transfer, for example in collisions, will be the same in all frames.

vanhees71
You are correct, it is not invariant.

Confidence restored!
You are correct. Consider the trivial case of a particle at rest in one frame. What is its kinetic energy in that frame, compared to in any other frame?

Energy transfer, for example in collisions, will be the same in all frames.
That's what I thought. Thanks

Isn't there an implicit assumption that the relationship ##E=\frac{p^{2}}{2m}## holds in all inertial frames though (or is it simply a matter of definition)?!

@Dale Whilst we're on the topic of Galilean invariance... I know that the instantaneous distance between two points is Galilean invariant, however if one considers a time interval, such that ##\Delta t\neq0##, then it is not. Is it correct to say though that it is meaningless to consider the distance between two points separated by a non-zero time interval since they are not defined in the same 3-dimensional hypersurface (with the set of hypersurfaces parametrised by absolute time ##t##), indeed they "live" on separate hypersurfaces?!

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vanhees71
Gold Member
2019 Award
Galileo transformations don't change time at all. So in fact time intervals are Galilei invariant. That's what Newton calls "absolute time".

You are right concerning spatial distances. Within Newtionian physics they make indeed only sense when looking at the two points defining the distance at the same time. There's no connection between the spaces at different times whatsoever. Newtonian spacetime is just a pile of 3D Euclidean affine spaces along the time axis (it's a fiber bundle rather than the elegant spacetime manifolds in special or general relativity).

alileo transformations don't change time at all. So in fact time intervals are Galilei invariant. That's what Newton calls "absolute time".

You are right concerning spatial distances. Within Newtionian physics they make indeed only sense when looking at the two points defining the distance at the same time. There's no connection between the spaces at different times whatsoever. Newtonian spacetime is just a pile of 3D Euclidean affine spaces along the time axis (it's a fiber bundle rather than the elegant spacetime manifolds in special or general relativity).
Is it assumed from the start that all equations (in Newtonian mechanics) are Galilean covariant (i.e. the are form invariant with respect to Galilean transformations)?! It is then a matter of showing that certain measurable quantities, such as Newton's 2nd law, are Galilean invariant, whereas others such as energy, momentum etc. are not.

vanhees71
Gold Member
2019 Award
Any observable (function of ##\vec{x}_j## and ##\vec{p}_j##, ##j \in \{1,2,\ldots,N \}## making up the ##3N##-dimensional phase space for a ##N##-body system) has a well-defined behavior under Galileo transformations. For Galileo boosts (a subgroup of the full Galileo group which is built by space-time translations, rotations, and Galileo boosts) you have
$$t'=t, \quad \vec{x}_j'=\vec{x}_j-\vec{v} t, \quad \vec{p}_j'=\vec{p}_j-m_j \vec{v}.$$
Now kinetic energy is
$$T'=\sum_{j=1}^N \frac{\vec{p}_j^{\prime 2}}{2m_j}=\sum_{j=1}^{N} \frac{(\vec{p}_j-m_j \vec{v})^2}{2m_j},$$
i.e., it's not invariant.

Any observable (function of ##\vec{x}_j## and ##\vec{p}_j##, ##j \in \{1,2,\ldots,N \}## making up the ##3N##-dimensional phase space for a ##N##-body system) has a well-defined behavior under Galileo transformations. For Galileo boosts (a subgroup of the full Galileo group which is built by space-time translations, rotations, and Galileo boosts) you have
$$t'=t, \quad \vec{x}_j'=\vec{x}_j-\vec{v} t, \quad \vec{p}_j'=\vec{p}_j-m_j \vec{v}.$$
Now kinetic energy is
$$T'=\sum_{j=1}^N \frac{\vec{p}_j^{\prime 2}}{2m_j}=\sum_{j=1}^{N} \frac{(\vec{p}_j-m_j \vec{v})^2}{2m_j},$$
i.e., it's not invariant.
Is it necessarily true that ##\mathbf{F}\rightarrow\mathbf{F}'##? Is this just a matter of definition? I understand that the right-hand side transforms as ##m\mathbf{a}\rightarrow m\mathbf{a}'##, but is it necessarily true that the left-hand side transforms such that it is equal to the transformed right-hand side?

vanhees71
Gold Member
2019 Award
A symmetry by definition means that the variation of the action doesn't change under the corresponding transformation, which implies that the equations of motion are invariant under the transformation. Since in the boost written out above ##\vec{v}=\text{const}## (i.e., ##\dot{\vec{v}}=0##) we have ##\dot{\vec{p}}_j'=\dot{\vec{p}}_j## any Galileo invariant force must be invariant itself. This gives constraints for the form of the possible Hamiltonians and thus the form of possible forces.

A symmetry by definition means that the variation of the action doesn't change under the corresponding transformation, which implies that the equations of motion are invariant under the transformation. Since in the boost written out above ##\vec{v}=\text{const}## (i.e., ##\dot{\vec{v}}=0##) we have ##\dot{\vec{p}}_j'=\dot{\vec{p}}_j## any Galileo invariant force must be invariant itself. This gives constraints for the form of the possible Hamiltonians and thus the form of possible forces.
So do we simply demand from the outset that all equations transform covariantly?

vanhees71
Gold Member
2019 Award
Yes, on a fundamental level the laws of nature should obey the symmetry principles of the underlying spacetime manifold.

Yes, on a fundamental level the laws of nature should obey the symmetry principles of the underlying spacetime manifold.
So in Newtonian mechanics the spacetime manifold has Galilean symmetry and so all equations should transform covariantly under this symmetry, meaning that, for example, ##\mathbf{F}\rightarrow\mathbf{F}'=m\mathbf{a}'##, ##\mathbf{p}\rightarrow\mathbf{p}'=m\mathbf{v}'##, ##E\rightarrow E'=\frac{\mathbf{p}'^{2}}{2m}+V'(\mathbf{x}')##, etc... And then it can shown that certain physical quantities are actually invariant under Galilean transformations, such as force, ##\mathbf{F}\rightarrow\mathbf{F}'=m\mathbf{a}'=m\mathbf{a}=\mathbf{F}##?!

vanhees71