How does Newton's Law change with rotation between frames?

[kent davidge]

If we have two frames related by $x' = Rx$ where $R$ is a rotation matrix and $t'=t$ Newton's law doesn't remain the same, for $$m \frac{d^2 x'}{dt'^2} = m \frac{d^2 Rx}{dt^2} = mRa$$ whereas it will be just $ma$ in the other frame. How do we solve this?

 

[kent davidge]

Just to give context to my question, I came to this after reading that a rotation is a possible Galilean transformation. See https://en.wikipedia.org/wiki/Galilean_transformation#Galilean_transformations

 

[bolometer]

Hi
It doesn't sound so bad to me, the $R$ matrix will rotate the $a$ vector because you are changing coordinates from the $S$ to the $S'$ reference frames.

 

[kent davidge]

Hi
It doesn't sound so bad to me, the $R$ matrix will rotate the $a$ vector because you are changing coordinates from the $S$ to the $S'$ reference frames.

Yea, but the force will not be the same in the two frames then

 

[Ibix]

The point is that $$F'=m\frac{d^2x'}{dt'}=m\frac{d^2(Rx)}{dt}=Rm\frac{d^2x}{dt}=RF$$In other words, if you rotate the coordinates then you rotate the representation of the force. Think about a force F in the x direction. What is F'?

 

[kent davidge]

The point is that $$F'=m\frac{d^2x'}{dt'}=m\frac{d^2(Rx)}{dt}=Rm\frac{d^2x}{dt}=RF$$In other words, if you rotate the coordinates then you rotate the representation of the force. Think about a force F in the x direction. What is F'?

Yes, I understand that

But the Galilean transformations are supposed to keep the force the same, that is $F' = F$. If we have $F' = RF$ this condition is not satisfied, despite the inner product being invariant as is required for an "orthogonal rotation".

Or... wait.. perhaps I have got things wrong and the magnitude of the force is required to be invariant under Galilean Transformations instead of the force itself. If that's the case then it's all right.

 

[Ibix]

Yes, I understand that

But the Galilean transformations are supposed to keep the force the same, that is $F' = F$. If we have $F' = RF$ this condition is not satisfied, despite the inner product being invariant as is required for an "orthogonal rotation".

Or... wait.. perhaps I have got things wrong and the magnitude of the force is required to be invariant under Galilean Transformations instead of the force itself. If that's the case then it's all right.

The force is the same. Its description in the two coordinate systems is different. Think about it like this - you and I are sat on opposite sides of a table. I finish my drink and push my glass into the middle of the table.

The force I applied is the same for you and me. It can't be different - there's only one physical situation. But from my perspective and yours the sign of the force is opposite. That's what's happening here.

 

[kent davidge]

Think about it like this - you and I are sat on opposite sides of a table. I finish my drink and push my glass into the middle of the table.

The force I applied is the same for you and me. It can't be different - there's only one physical situation. But from my perspective and yours the sign of the force is opposite. That's what's happening here.

Good illustration. But then isn't it more correct to say that the magnitude of the force is the same for you and me? Instead of just the force..

 

[Nugatory]

Good illustration. But then isn't it more correct to say that the magnitude of the force is the same for you and me? Instead of just the force..

No, the force, which is a vector, is the same. The components of that vector will change when you change coordinate systems, but that's just a different way of writing the same vector.

 

[kent davidge]

No, the force, which is a vector, is the same

But then how do we deal with the example posed by @Ibix of one force having negative sign respective to the other force?

 

[Ibix]

But then how do we deal with the example posed by @Ibix of one force having negative sign respective to the other force?

The force is what it is. It's a physical thing. We are both coming up with a description of the same thing using different schemes.

Going back to the table - suppose it's rectangular. According to you and me, sat opposite each other, it's long and narrow. According to someone sat on the side, though, it's short and wide. It's the same table either way. Similarly, it's still the same force whether we choose to describe it relative to your coordinate system or mine.

 

[Nugatory]

But then how do we deal with the example posed by @Ibix of one force having negative sign respective to the other force?

You and ibix are using different coordinate systems. The transformation between them is $x'=-x$, $y'=y$, $z'=z$, $t'=t$ (assuming that the x-axis is chosen so that the glass is moving along it).

 

[kent davidge]

Hmm, ok. But then we can perform whatever coordinate transformation we like, not just those Galilean transformations, correct?

 

[Ibix]

Yes. But in general, this will cause extra terms to appear in equations - so called "fictitious forces".

 

[kent davidge]

Yes. But in general, this will cause extra terms to appear in equations - so called "fictitious forces".

That's the case for a constant rotation matrix that I talked bout in the opening post. So does this mean that even Galilean transformations can give rise to such extra terms?

 

[Nugatory]

That's the case for a constant rotation matrix that I talked bout in the opening post. So does this mean that even Galilean transformations can give rise to such extra terms?

That constant rotation doesn't introduce any fictitious forces, it just changes the direction of the axes. Only if you make the rotation angle a non-constant function of time (for example, coordinates in which an observer on a merry-go-round is at rest while the amusement park is rotating) will the fictitious centrifugal force appear.

Some of the problem here may be that in your original post you misapplied some special-case assumptions; for example $a=\frac{d^2x}{dt^2}$ only works if the force and velocity are both vectors parallel to the x-axis, and clearly if the x and x' axes are related by a rotation that can't be the case in both coordinate systems. Instead, you have to use the more general approach: write the coordinates of the position of the object as functions of time: $x(t)$, $y(t)$, $z(t)$. (We generally make the notation more compact by writing these three expressions as $x_i(t)$, so I'll use that convention from here on).

Now we have $v_i=\frac{dx_i}{dt}$ and $a_i=\frac{d^2x_i}{dt^2}$ (and of course $\vec{F}=m\vec{a}$ where the $a_i$ are the components of $\vec{a}$). Transform the three $x_i$ to the corresponding $x'_i$ and you'll find that these relations and Newton's second law work just fine in the primed frame.

 

[Dale]

If we have two frames related by $x' = Rx$ where $R$ is a rotation matrix and $t'=t$ Newton's law doesn't remain the same, for $$m \frac{d^2 x'}{dt'^2} = m \frac{d^2 Rx}{dt^2} = mRa$$ whereas it will be just $ma$ in the other frame. How do we solve this?

Don’t forget $a’=Ra$ so

$$m \frac{d^2 x'}{dt'^2} = m \frac{d^2 Rx}{dt^2} = mRa = ma’$$

 

[Dale]

Good illustration. But then isn't it more correct to say that the magnitude of the force is the same for you and me? Instead of just the force..

It is correct to say that it is the force that is the same, not just the magnitude of the force. What is changing is the coordinate basis, not the vector itself. The vector itself remains the same.

For instance, suppose we have the rotation
$\hat{x’}=\hat{y}$
$\hat{y’}=-\hat{x}$
$\hat{z’}=\hat{z}$
Where the hats denote the orthonormal basis vectors.

Then the vector
$\vec{f}=a \hat{x} = -a \hat{y} = \vec{f’}$

They are in fact the same vector even though in the coordinate bases $\vec{f}=(a,0,0)$ and $\vec{f’}=(0,-a,0)$

 

[Ibix]

That's the case for a constant rotation matrix that I talked bout in the opening post. So does this mean that even Galilean transformations can give rise to such extra terms?

No it's not. As Nugatory and Dale pointed out, it's just a change of description. In one frame you have $F=ma$, in the other you have $F'=ma'$, where $F'=RF$ and $a'=Ra$.

But if you have (e.g.) a rotating frame then your R is a function of time and as well as an $Rd^2\vec {x}/dt^2$ term the chain rule gives you an $\vec{x}d^2R/dt^2$ term. The first term tells you your coordinates are different, but the second is present even when there's no acceleration. It's why it feels like you are being pressed against the door when you go round a corner in a car.

 

[kent davidge]

Thanks to all. I got it.