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Thinking this would be obvious, I was trying to prove the Galilean invariance of Newton's second law of motion, but I failed. This is what I've got so far:

If we define a world line in Galilean space-time $\mathbb{R}^{4}$ as the following curve $$\bar{w}\colon I\subset\mathbb{R}\to \mathbb{R}^{4}\colon t\mapsto (t,\bar{x}(t))$$ $$\bar{x}\colon I\subset\mathbb{R}\to \mathbb{R}^{3}\colon t\mapsto (x(t),y(t),z(t))$$

where $\mathbb{R}^{3}\subset\mathbb{R}^{4}$ Euclidean, then the acceleration is given by

$$\bar{a}\colon I\subset\mathbb{R}\to \mathbb{R}^{4}\colon t\mapsto \frac{d^{2}\bar{w}(t)}{dt^{2}}=(0,\frac{d^{2}\bar{x}(t)}{dt^{2}})\equiv(0,\tilde{a}(t))$$

where $\tilde{a}$ the classical acceleration and the force field that causes the acceleration

$$\bar{F}\colon\mathbb{R}^{4}\to\mathbb{R}^{4}\colon\bar{w}(t)\mapsto m\bar{a}(t)=(0,m\tilde{a}(t))$$

So if Newton's second law of motion is written as $\bar{F}(\bar{w}(t))=m\bar{a}(t)$ then a Galilean transformation causes $\bar{F}=G\bar{F}'$ and $\bar{a}=G\bar{a}'$ since they both live in $\mathbb{R}^{4}$. Therefore $$\bar{F}(\bar{w}(t))=m\bar{a}(t)$$ $$\Leftrightarrow G\bar{F}'(\bar{w}(t))=mG\bar{a}'(t)$$ $$\Leftrightarrow \bar{F}'(\bar{w}(t))=m\bar{a}'(t)$$ which is what we needed to prove (if F=ma in the unprimed frame then F'=ma' in the primed frame). Note that this is analogue to how Lorentz invariance is shown in special relativity.

However $\bar{a}$ doesn't transform like $\bar{a}=G\bar{a}'$

A general Galilean transformation $G$ in $\mathbb{R}^{4}$ is given by $$t=t'+t_{t}$$ $$\bar{x}=R\bar{x}'+\bar{u}t+\bar{t}_{\bar{x}}$$

The relation between velocity and acceleration before and after a Galilean transformation is given by $$\bar{v}(t)=\frac{d\bar{w}(t)}{dt}=\frac{d\bar{w}(t)}{dt'}\frac{dt'}{dt}=\frac{d\bar{w}(t)}{dt'}= (1,R\frac{d\bar{x}'}{dt'}+\bar{u})$$ $$\bar{a}(t)=(0,\tilde{a}(t))=\frac{d\bar{v}(t)}{dt}=(0,R\frac{d^{2}\bar{x}'}{dt'^{2}})=(0,R\tilde{a}'(t'))$$ $$\Leftrightarrow\tilde{a}(t)=R\tilde{a}'(t')$$

This is not the same as $\bar{a}=G\bar{a}'$ because

$$\bar{a}(t)=G\bar{a}'(t')$$ $$\Leftrightarrow\begin{pmatrix} 0 \\ \tilde{a}(t) \\ 1\end{pmatrix} = \begin{pmatrix}1&0&t_{t}\\ \bar{u}&R&\bar{t}_{\bar{x}}\\ 0&0&1 \end{pmatrix}\cdot\begin{pmatrix} 0 \\ \tilde{a}'(t') \\ 1\end{pmatrix}$$

So the acceleration of a world line transforms not with $G$ but with the linear part of $G$, meaning that the translation part must be zero: $(t_{t},\bar{t}_{\bar{x}})=\bar{0}$.

Can someone help me out of this mess?

Edit: Let me try again, this time forgetting that we're talking about forces and just consider a 3D vector field. Of course Galilean space-time is still 4-dimensional and a general Galilean transformation $G$ in $\mathbb{R}^{4}$ is still given by $$t=t'+t_{t}$$ $$\bar{x}=R\bar{x}'+\bar{u}t+\bar{t}_{\bar{x}}$$ and the relation between the classical acceleration in inertial frames (primed and unprimed) related by a Galilean transformation $G$ is still given by $$\tilde{a}(t)=R\tilde{a}'(t')$$ One could say that the acceleration transforms with $R$ if the frame transforms with $G$ because $\tilde{a}$ lives in the associated vector space of Galilean space time (therefore the affine translation $(t_{t},\bar{t}_{\bar{x}})$ doesn't apply) and moreover lives in the 3D Euclidean subspace of this vector space $\mathbb{R}^{3}\subset\mathbb{R}^{4}$ (therefore the boost $\bar{u}$ doesn't apply).

Suppose now that we define a 3D vector field on a world line as $$\bar{F}\colon C\subset\mathbb{R}^{4}\to\mathbb{R}^{3}\colon\bar{w}(t)\mapsto m\tilde{a}(t)$$ In this case, by the same reasoning as for the acceleration, we can say that the vector field (defined in the unprimed frame) transforms with $R$ if the frame transforms with $G$ $$\bar{F}(\bar{w}(t))=R\bar{F}'(\bar{w}'(t'))$$ If we use this together with $$\bar{F}(\bar{w}(t))=m\tilde{a}(t)=mR\tilde{a}'(t')$$ it follows that $$\bar{F}'(\bar{w}'(t'))=m\tilde{a}'(t')$$ So if we define a vector field as $\bar{F}(\bar{w}(t))=m\tilde{a}(t)$ (forget that we're talking about force) and since both sides live in the same space ($\mathbb{R}^{3}$), then they also transform in the same way under a Galilean transformation (whatever this way is, in this case $R$). Therefore it doesn't matter in which frame we define the vector field on a world line as the acceleration multiplied by the mass, it will have the same form in all frames. Therefore we could say that the definition of the vector field is Galilean invariant.

The problem I'm having is that one says that "Newton's second law of motion is Galilean invariant". This implies that it is invariant, regardless the nature of the force. So if we make the nature of the force abstract, we can just forget that we're talking about force and consider a 3D vector field. Then everything boils down to showing that the definition has the same form in all inertial frames, which it has as shown above. Is this a valid point of view?

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The problem is the maths - it's covering up the physics, making it difficult to see what you're doing. –  Larry Harson Jun 1 '12 at 13:24
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It's a real mess. What does it have to do with the simple problem of Galilean invariance? Just replace all $\vec x\mapsto \vec x+ \vec V t$ and $\vec v_i\mapsto \vec v_i+\vec V$ and check that Newton's equations hold if they held before. That's simple because the relative distances won't change and the dependence on derivatives is only on accelerations where $\vec V$ cancels. –  Luboš Motl Jun 1 '12 at 13:26
    
Everyone keeps telling me this is simple, but when I ask them to write down the actual proof (not just the outline of a proof), they fail. So can you show me how you would prove the Galilean invariance of F=m.a? Also use the general Galilean transformation and not the special case you refer to (many books do that, I know). –  Wox Jun 1 '12 at 14:27
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Your mistake is simply that you assume that $F$ and $a$ have to transform according to $G$ under Galilean transformations. That is not true however. $F$ and $a$ do not change under translation and boosts. Only under rotations. –  Raskolnikov Jun 1 '12 at 15:05
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@Wox, Luboš Motl's answer is correct. Formulate F=ma as F=mx'' where $x=x_0+v_0 t+ \frac{1}{2} at^2$ Then $x''=a$. Now apply the Galilean Transformation to this.$\bar{x}=x_0+v_0 t+ \frac{1}{2}at^2 +Vt$ and notice that again $\bar{x}''=a$ –  MadScientist Jun 1 '12 at 15:41
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1 Answer 1

The question is relevant. The answer is not simple. In fact, the acceleration is a special geometrical object, distinct of a point or a vector. It is transformed as an affine connection (see for instance in wikipedia or in differencial geometry textbooks). In particular, it is worth to consider the principal connections, invariant in a certain sense with respect to a group action.

In General Relativity, the suitable group is the Poincaré group. The linear transformations of this group are Lorentz transformations.

In the Galilean Relativity, it is Galileo group, that leads to consider the Galilean connections (or Newton-Cartan connections).

For more details, you can read for instance: G. de Saxce, C. Vallee, "Affine tensors in shell theory", Journal of theoretical and applied mechanics, 41(3), pp 593-621 (2003).

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protected by Qmechanic Mar 18 '13 at 23:04

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