I Force on Scale in Rocket Scenario: Calculation & Discussion

[PeterDonis]

Calculate the invariant: the proper acceleration of the object on the scale, times the object's invariant mass.

To correct this based on @pervect's posts: I had (incorrectly) thought that the block's proper acceleration was the same as the rocket's, but as the previous thread referenced by @pervect makes clear, it is larger by a factor of $\gamma^2$. So a scale attached to the block will read $\gamma^2 m a$, where $a$ is the proper acceleration of the rocket. @pervect's discussion of the length contraction of the contact area then explains why the scale attached to the rocket (instead of the block) only reads $\gamma m a$.

 

[Sagittarius A-Star]

The pressure / unit area under the sliding block will be $T^{zz} = T_{zz} = \gamma^2 \rho \, A$, $\rho$ being the density of the block, and A being the vertical acceleration of the rocket floor.

As I understand, the factor $\gamma^2$ means, that the proper acceleration of the block is by this factor greater than the proper acceleration of the scale.

The four-acceleration of the block is
$\mathbf A = \begin{pmatrix} \gamma \dot \gamma c \\ \gamma^2 \vec a + \gamma \dot \gamma \vec v \\ \end{pmatrix}$
The block's $\dot \gamma = \frac{\vec a \cdot \vec v}{c^2}\gamma^3$.
Source:
https://en.wikipedia.org/wiki/Four-acceleration

Because $\vec a$ is vertical and $\vec v$ is horizontal, $\dot \gamma =0$. That means:
$\mathbf A = \begin{pmatrix} 0 \\ \gamma^2 \vec a \\ \end{pmatrix}$

=> The proper acceleration of the block is $\alpha = \gamma^2 a$.

Reason:

Therefore, the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration that a moving particle "feels" moving along a worldline.

Source:
https://en.wikipedia.org/wiki/Four-acceleration

The four-force from the block is
$\mathbf F = \begin{pmatrix} \gamma \frac{\vec f \cdot \vec u}{c} \\ \gamma \vec f \end{pmatrix} = m \mathbf A$
Source:
https://en.wikipedia.org/wiki/Four-force
=> $\gamma \vec f = m \gamma^2 \vec a$

=> The force in the reference frame is $\vec f = m \gamma \vec a$.

 

[PeterDonis]

As I understand, the factor $\gamma^2$ means, that the proper acceleration of the block is by this factor greater than the proper acceleration of the scale.

If the scale is fixed to the rocket, not the block, yes.

=> The force in the reference frame is $\vec f = m \gamma \vec a$.

No, the force measured by the scale fixed to the rocket is $m \gamma \vec{a}$. This measured force is an invariant; it is the same no matter what frame you use to calculate it.

You appear to be confusing "reference frame" with "state of motion of a physical object". The reading on the scale fixed to the rocket, registered by the block moving transversely relative to the rocket, is not a matter of different "reference frames". It is a matter of the different states of motion of the scale and the block. That difference is a physical fact, independent of any choice of frame.

 

[pervect]

=> The proper acceleration of the block is $\alpha = \gamma^2 a$.

I concur, though I used a different method to compute the same result.

The four-force from the block is

The force is not at a point, but it is distributed over some area. The area the force is distributed over is not the same in the rocket frame and the block frame. This happens because of the relativity of simultaneity. To get the total force, one has to integrate the force/unit area over the contact area at some particular "instant of time". Because of the relativity of simlutaneity, the concept of "an instant of time" is frame dependent. As a consequence, the contact area at some particular "instant of time" is also frame dependent, it is different in the block frame and in the rocket frame.

To accommodate the fact that the force is distributed, rather than occurring at a point (as is the case for a four-force), one can use the notion of the stress-energy tensor. Certain components of this tensor mathematically model force/unit area. These are the "pressure" terms in the stress-energy tensor. This may not make much sense if you are not familiar with the stress energy tensor, I suppose.

The stress-energy tensor components transform in a known manner. In the limit of a thin block, we can use the Lorentz transform to do this transformation. The stress-energy tensor is not a vector (a rank 1 tensor), but rather is a rank 2 tensor. It turns out, though, that the vertical component of the pressure in the SET is the same in both the rocket frame and the block frame, because of the appropriate transformation laws, which is rather convenient.

The end result of this process is the same result you get, it's just a different (and IMO better) way of looking at what's happening. It's better because the SET models a distribute force. However, some familiarity with the stress-energy tensor is required for this viewpoint to be useful, I suppose.