Force on Scale in Rocket Scenario: Calculation & Discussion

[Sagittarius A-Star]

TL;DR Summary

What is the force on a scale in an accelerated rocket in flat spacetime in the following scenario (SR)?

[Moderator's note: Spin-off posts from previous thread have been included in this new thread. Also, the OP's re-post of the scenario for discussion has been moved to this top post for clarity.]

The downward proper acceleration of an object on the surface of a planet like the Earth is independent of tangential velocity

Yes.

So scales don't measure relativistic mass anyway, they measure invariant mass

Physically, scales measure a force (and indirectly the energy) in their frame. Consider the same scenario in a large accelerating elevator cabin in space and describe it in an inertial frame, in which $v_z/c$ is small for a certain time ($v_x$ is relativistic and constant).

$dp_z = m\gamma dv_z$
$a_z = dv_z/dt$
$F_z= dp_z/dt$
$F_z/a_z = dp_z/dv_z = m\gamma$

Source:
http://www.physikdidaktik.uni-karlsruhe.de/download/195_longitudinal_and_transverse_mass.pdf

[Moderator's note: Including later re-post of the scenario here.]

A rocket in space has at it's stern a constant proper acceleration of 9.81 m/s² (in z-direction).

On the floor at the stern of the rocket stands a scale. A toy-car with invariant mass $m$ drives with constant velocity $v_x$ over the scale. What is the force on the scale?

As reference frame for the calculation, I choose a co-moving inertial frame, which is at rest relative to the accelerating rocket at an instance of time, in which I do the calculation. I make use of the following for the inertial reference frame:

it is always an IF, it always moves uniformly: an accelerating particle co-moves with its rest-frame for only one instant. But at that instant the rest-frame measures the particle's proper acceleration.

Source:
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Four-force_and_three-force

So in this inertial frame - at the described instance of time - the car moves in x-direction with $v$ and accelerates together with the rocket-stern in z-direction with 9.81 m/s².

$dp_z = m\gamma dv_z$
$a_z = dv_z/dt$
$F_z= dp_z/dt$
$F_z/a_z = dp_z/dv_z = m\gamma$

=> The force on the scale is $F_z = m\gamma \cdot 9.81 m/s^2$.

Of course, the ratio $F_z/a_z$ in this scenario should not been called "traverse mass" nor "relativistic mass", because that would only create confusion regarding the word "mass".

Is my above calculation correct and if not, why exactly?
What is a possible better way to calculate the force on the scale?

 

[vanhees71]

Well, the Karlsruhe physics course... It's considered to be in Pauli's category of "not even wrong".

One should note that gravitation has to be treated within general relativity. I'd keep this out of this thread completely, because the OP doesn't even know the scientific method to a sufficient degree to understand special relativity.

For any other interaction than gravity, the most simple model for the motion of a point particle is the motion in some external field. The most simple (and practically most important) example is the motion of a charged particle in an electromagnetic field, which is described by an antisymmetric tensor whose components are the electric and magnetic field components $F_{0j}=E^j$ and $F_{jk}=-\epsilon^{jkl}B^l$.

Then the manifestly covariant equation of motion reads
$$m \ddot{x}^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu},$$
where the dot means differentiation wrt. proper time. Of the four equations only three (you can choose the spatial components) are independent, because of the constraint
$$\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}.$$
Again, of course, $m=\text{const}$, is the invariant mass of the particle. No complicated non-covariant ideas from the first few years of SR are necessary, particularly no transverse and longitudinal relativistic masses.

 

[Sagittarius A-Star]

The most simple (and practically most important) example is the motion of a charged particle in an electromagnetic field,

In this case you get into $F/a$ in transversal direction the $\gamma$, too:
https://www.mathpages.com/home/kmath674/kmath674.htm

Also, consider in an accelerating rocket in space two toy-cars, each with mass $m$, driving with the same relativistic speed in opposite direction over a scale. The mass of the system is $M = 2m\gamma$. Does the scale measure $M$?

 

[vanhees71]

I gave the precise equations of motion. You can reformulate them in the (1+3) non-manifestly covariant form, which is completely equivalent, but it doesn't make the slightest sense from any point of view to introduce the overcomplicated idea of several kinds of "relativistic mass".

For composite systems by definition the total mass is the energy in the rest frame of the center-momentum frame divided by $c^2$. As I said, I'd keep out gravity from the discussion at this level completely.

 

[Sagittarius A-Star]

non-manifestly covariant form, which is completely equivalent, but it doesn't make the slightest sense from any point of view to introduce the overcomplicated idea of several kinds of "relativistic mass".

I agree. Therefore I spoke always of "ratio $F/a$" (which I need to understand the scale measurement) or "energy" (not of "relativistic mass"), and the system mass $M$, I mentioned before, is invariant.

I said, I'd keep out gravity from the discussion at this level completely.

Therefore, I discussed the accelerating rocket within SR.

 

[PeterDonis]

Physically, scales measure a force

Actually, most scales directly measure a displacement; the force is calculated from the spring constant of the spring inside the scale.

(and indirectly the energy)

I'm not sure what you mean. There is no direct connection between the force measured by the scale and any energy except for the potential energy stored in the spring.

in their frame.

No, the measurement given by a scale, like any direct observable, is invariant; it's the same no matter what frame you adopt.

Consider the same scenario in a large accelerating elevator cabin in space and describe it in an inertial frame

The quantity you are calculating is not the invariant that the scale measures. The force the scale measures the proper acceleration of the object times its invariant mass.

 

[Sagittarius A-Star]

The quantity you are calculating is not the invariant that the scale measures.

Why does the scale measure an invariant? Please consider also:

Note that the transformed force in general depends not only on the original force but also on the velocity $u$ of the particle on which the force acts. Thus a velocity-independent force (like Newton's gravitational force field) is no longer a Lorentz-invariant concept.

Source:
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Four-force_and_three-force

 

[Ibix]

Why does the scale measure an invariant?

All measurements are invariants. Otherwise I could use two different frames to get two different predictions of what the measurement (an actual number on an actual machine) would be. Generally, if you think you are measuring a component of a vector you are measuring the inner product of that vector and the relevant basis vector.

 

[Sagittarius A-Star]

All measurements are invariants.

Does it mean that the following spring measures the invariant mass?

To expand on this a bit, apply the equivalence principle and consider a rocket accelerating in flat spacetime. Inside you have a light strong frictionless horizontal rail suspended from a spring, and a small body of mass $m$ moving with (as measured in this frame) constant velocity along the rail. How much force would the spring exert in equilibrium? Since the spring's force is perpendicular to the direction of motion the answer is $\gamma ma$, where $a$ is the "acceleration due to gravity".

 

[Ibix]

Does it mean that the following spring measures the invariant mass?

No. It means that it measures the inner product of the four force with a four vector along the length of the spring. That's a Lorentz scalar and therefore invariant, but it isn't the invariant mass.

 

[PeroK]

Does it mean that the following spring measures the invariant mass?

IMO your confusion is a prime example of the confusion caused by using relativistic mass.

There is only one number showing on the scales. How you interpret that number may be flexible.

This is the fundamental problem. There is no way to say that ultimately a certain contraction of a spring is a measure of invariant mass or relativistic mass.

 

[Sagittarius A-Star]

There is no way to say that ultimately a certain contraction of a spring is a measure of invariant mass or relativistic mass.

So scales don't measure relativistic mass anyway, they measure invariant mass

Since the spring's force is perpendicular to the direction of motion the answer is $\gamma ma$, where $a$ is the "acceleration due to gravity".

What of the three is correct?

 

[Sagittarius A-Star]

... with a four vector along the length of the spring.

What kind of four-vector is this?

 

[vanhees71]

You should concisely specify how you want to measure mass or energy or whatever with your spring. Only then one can answer what's measured.

 

[PeroK]

What of the three is correct?

You need to stop flogging this dead horse.

Relativistic mass is a convention where you multiply the invariant mass by the gamma factor. The distinction is not something physically measurable, as you always have these two quantities in the equation somewhere.

 

[Sagittarius A-Star]

You need to stop flogging this dead horse.

Relativistic mass is a convention where you multiply the invariant mass by the gamma factor. The distinction is not something physically measurable, as you always have these two quantities in the equation somewhere.

I agree that the scale measures the force. I always use $m$ for the "invariant mass" and not for the "relativistic mass".

 

[PeterDonis]

Why does the scale measure an invariant?

Because any measured quantity must be an invariant. What you read on a scale cannot depend on what coordinates you choose.

 

[PeterDonis]

Is my above calculation correct

No.

if not, why exactly?

Because it's calculating a coordinate-dependent quantity, and no actual measurement result can be a coordinate-dependent quantity. The reading displayed on the scale cannot depend on what coordinates you choose.

What is a possible better way to calculate the force on the scale?

Calculate the invariant: the proper acceleration of the object on the scale, times the object's invariant mass. You specified both of those things in your problem statement, so the calculation is a one-liner.

 

[Sagittarius A-Star]

Because it's calculating a coordinate-dependent quantity, and no actual measurement result can be a coordinate-dependent quantity. The reading displayed on the scale cannot depend on what coordinates you choose.

The measurement-result is invariant, yes. But that does not mean, that the measured quantity must also be an invariant (see for example speed-tickets from the police).

Calculate the invariant: the proper acceleration of the object on the scale, times the object's invariant mass. You specified both of those things in your problem statement, so the calculation is a one-liner.

Wouldn't that be the force in the rest-frame of the car and not in the specified reference-frame?

Forces and spring-constants are frame-dependent, as shown in the following example:
https://www.mathpages.com/home/kmath068/kmath068.htm

In the rest-frame of the car, the spring-constant (and therefore the measurement-result) depends on the velocity $-v$ of the scale.

 

[Ibix]

But that does not mean, that the measured quantity must also be an invariant (see for example speed-tickets from the police).

Yes it does - your speed as measured by a policeman is the arccosh of the inner product of your four velocity and the cop's. This is a Lorentz scalar and an invariant. No other frame will interpret that particular invariant as "your speed", but all will agree the value.

 

[Ibix]

All have to agree the value the cop measures, because otherwise they cannot explain why his Doppler radar gun displays the value it does, and there's nothing stopping them reading the display as they pass.

 

[pervect]

I haven't been following this at all, but the following remark caught my eye. I have to disagree with it, having analyzed similar problems in the past.

The downward proper acceleration of an object on the surface of a planet like the Earth is independent of tangential velocity

The downward proper acceleration on an object on a planet like the Earth is not independent of the tangential velocity. For instance, consider an object orbiting the Earth at zero altitude, whose proper acceleration is zero.

Part of this is because the Earth's surface is curved, of course. But if we ask the question, is the downward proper acceleration of an object on the floor of Einstein's elevator independent of tangential velocity, the answer is still no.

This goes back to the rather old "sliding block" thread. but we had a more recent discussion in https://www.physicsforums.com/threa...ng-at-relativistic-speeds.948690/post-6008592

where I consider the proper acceleration of an object at rest on the floor of Einstein's elevator, and another object moving transversely across the floor of Einstein's elevator. The result in the coordinates used (see the post for details, the coordinates are implied by the line element in that post) is $Z / (Z^2 - \beta^2)$ , in the geometric units used in that post. In non-geometric units it'd be $c^2 \,Z/ (Z^2 - \beta^2)$.

Here $\beta$ can be regarded the normalized velocity v/c of the object relative to the floor of the elevator. When we set Z=1 which is the Z coordinate of the "ground level" of the elevator in the coordinates used, this works out to be $1 / (1-\beta^2)$ for the object moving at a velocity v, and 1 for the object with $v= \beta=0$.

I'd have to read more of the thread to comment on the original point, I hope to get to that later, but I noticed this straight off.

 

[Sagittarius A-Star]

Yes it does - your speed as measured by a policeman is the arccosh of the inner product of your four velocity and the cop's. This is a Lorentz scalar and an invariant. No other frame will interpret that particular invariant as "your speed", but all will agree the value.

I think, I used wrongly the expression "measured quantity". I did not mean with it "speed, as referenced to a certain frame", but speeds in general. And in general, a speed $v<c$ is frame-dependent, like for example $\frac{E}{c^2}$.

 

[Sagittarius A-Star]

All have to agree the value the cop measures

Surely not the person, who received the speed-ticket

 

[Ibix]

I think, I used wrongly the expression "measured quantity". I did not mean with it "speed, as referenced to a certain frame", but speeds in general. And in general, a speed $v<c$ is frame-dependent, like for example $\frac{E}{c^2}$.

But you can't measure "speeds in general". You can only measure speed with respect to some state of rest - in which case you're measuring the inner product of the four velocities of the object and something that you've defined as at rest.

 

[anuttarasammyak]

Summary:: What is the force on a scale in an accelerated rocket in flat spacetime in the following scenario (SR)?

=> The force on the scale is Fz=mγ⋅9.81m/s2.

The result seems reasonable to me because scale measures energy including kinetic one, on it, not mass or , in old words, rest mass.

 

[PeterDonis]

The measurement-result is invariant, yes. But that does not mean, that the measured quantity must also be an invariant

The measured quantity is the measured result.

see for example speed-tickets from the police

The reading on the policeman's radar gun is an invariant; it's the same no matter what coordinates you adopt.

What you appear to be thinking of is the reading on the policeman's radar gun, where he is sitting by the side of the road and you are going by at 150 kph, as compared to the reading on a different radar gun, carried by you in the car, which of course will read zero. These are two different invariants. They are not two different versions of the same "measured quantity" in two different frames. They are two different measured quantities, measured by two different devices in different states of motion.

Much of this thread is basically you not grasping the basic concepts I have just described.

 

[pervect]

Summary:: What is the force on a scale in an accelerated rocket in flat spacetime in the following scenario (SR)?
A rocket in space has at it's stern a constant proper acceleration of 9.81 m/s² (in z-direction).

On the floor at the stern of the rocket stands a scale. A toy-car with invariant mass $m$ drives with constant velocity $v_x$ over the scale. What is the force on the scale?

...

=> The force on the scale is $F_z = m\gamma \cdot 9.81 m/s^2$.

I concur. I prefer a slightly different formulation of the problem. Consider a flat sliding block, so that the contact area of the block is the entire surface of the block surface, rather than just the area under the wheels of the toy car.

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.

However, the contact area of the sliding block will be $1/\gamma$ due to length contraction. The the total force as measured by a truck scale on the rocket floor will be the pressure multiplied by the contact area, for the same result you find.

That's how the truck scale (a scale measuring the weight of a moving object) works. Physically, it reacts to the pressure of the object passing over it, and integrates the pressure over the contact area, with an appropriate and implied simultaneity convention, to find the force.

If we consider a similar scale on the sliding block, the pressure under it will be unchanged, as $T^zz$ does not change under a transverse boost. The contact area will be larger, though, and the scale will have a correspondingly greater reading in the block frame. I believe this is consistent with the transformation law of forces between frames, the difference being between dp/dt and $dp/d\tau$.

 

[PeterDonis]

scale measures energy including kinetic

There is a sense in which this is true, but it's not any sense that is relevant to this thread.

If I have a top on a scale, not rotating, and I measure its weight, and then I set the top spinning very rapidly, and measure its weight again, the weight will increase (by an unmeasurably tiny amount for any actual top and any actual spin I can impart with my hands, but in principle we could set up an experiment with a very sensitive scale and a mechanism to spin a top rapidly enough to detect the change). In that sense, yes, the rotational kinetic energy of the top contributes to the weight measured by the scale.

However, from the point of view of relativity, this rotational kinetic energy is part of the top's invariant mass. In other words, by spinning the top, I am increasing its invariant mass. I am not adding something besides invariant mass that gets measured as weight.

Of course we could adopt a more detailed model of the top, in which we modeled the invariant masses of each atom; then the kinetic energy of rotation would not increase the invariant masses of the individual atoms. But it still would increase the invariant mass of the top as a whole, because invariant mass is not additive; you can act on a system in a way that increases the system's invariant mass, without changing the invariant masses of any of its constituents.

 

[PeterDonis]

...if we ask the question, is the downward proper acceleration of an object on the floor of Einstein's elevator independent of tangential velocity, the answer is still no.

This goes back to the rather old "sliding block" thread. but we had a more recent discussion in https://www.physicsforums.com/threa...ng-at-relativistic-speeds.948690/post-6008592

Thanks for the link @pervect, I had dimly remembered this coming up before in a thread you were involved in but I evidently misremembered the details.

Looking at the more recent thread, I note that there are actually two possible things that can be measured. The proper acceleration of the sliding block, which would be measured by a scale moving with the block, is larger than the rocket's proper acceleration by a factor of $\gamma^2$. However, the measurement on a scale that is not sliding with the block is only larger by a factor of $\gamma$. The difference, as you note in your later post in this thread, can be attributed to the length contraction of the block in the rocket's instantaneous rest frame, which reduces its contact area with the rocket-fixed scale by a factor of $\gamma$.

 

[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.