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 R4 as the following curve ˉw:I⊂R→R4:t↦(t,ˉx(t)) ˉx:I⊂R→R3:t↦(x(t),y(t),z(t))
where R3⊂R4 Euclidean, then the acceleration is given by
ˉa:I⊂R→R4:t↦d2ˉw(t)dt2=(0,d2ˉx(t)dt2)≡(0,˜a(t))
where ˜a the classical acceleration and the force field that causes the acceleration
ˉF:R4→R4:ˉw(t)↦mˉa(t)=(0,m˜a(t))
So if Newton's second law of motion is written as ˉF(ˉw(t))=mˉa(t) then a Galilean transformation causes ˉF=GˉF′ and ˉa=Gˉa′ since they both live in R4. Therefore ˉF(ˉw(t))=mˉa(t) ⇔GˉF′(ˉw(t))=mGˉa′(t) ⇔ˉF′(ˉw(t))=mˉa′(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 ˉa doesn't transform like ˉa=Gˉa′
A general Galilean transformation G in R4 is given by t=t′+tt ˉx=Rˉx′+ˉut+ˉtˉx
The relation between velocity and acceleration before and after a Galilean transformation is given by ˉv(t)=dˉw(t)dt=dˉw(t)dt′dt′dt=dˉw(t)dt′=(1,Rdˉx′dt′+ˉu) ˉa(t)=(0,˜a(t))=dˉv(t)dt=(0,Rd2ˉx′dt′2)=(0,R˜a′(t′)) ⇔˜a(t)=R˜a′(t′)
This is not the same as ˉa=Gˉa′ because
ˉa(t)=Gˉa′(t′) ⇔(0˜a(t)1)=(10ttˉuRˉtˉx001)⋅(0˜a′(t′)1)
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: (tt,ˉtˉx)=ˉ0.
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 R4 is still given by t=t′+tt ˉx=Rˉx′+ˉut+ˉtˉx and the relation between the classical acceleration in inertial frames (primed and unprimed) related by a Galilean transformation G is still given by ˜a(t)=R˜a′(t′) One could say that the acceleration transforms with R if the frame transforms with G because ˜a lives in the associated vector space of Galilean space time (therefore the affine translation (tt,ˉtˉx) doesn't apply) and moreover lives in the 3D Euclidean subspace of this vector space R3⊂R4 (therefore the boost ˉu doesn't apply).
Suppose now that we define a 3D vector field on a world line as ˉF:C⊂R4→R3:ˉw(t)↦m˜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 ˉF(ˉw(t))=RˉF′(ˉw′(t′)) If we use this together with ˉF(ˉw(t))=m˜a(t)=mR˜a′(t′) it follows that ˉF′(ˉw′(t′))=m˜a′(t′) So if we define a vector field as ˉF(ˉw(t))=m˜a(t) (forget that we're talking about force) and since both sides live in the same space (R3), 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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