# Reading on a scale inside a train

1. Dec 29, 2013

### bcrowell

Staff Emeritus
I recently came across this paper http://arxiv.org/abs/physics/0504110v2 by Oas that is a semi-rant about how pedagogically evil relativistic mass is. He gives the following sample question:

From the context, it appears that the students are expected to have little or no background in relativity other than reading a paragraph that introduces relativistic mass in special relativity. It seems from the context that they do *not* know the equivalence principle (or anything about GR). The correct answer is stated to be that in both cases, a and b, the reading equals 200 lb. He takes the students' failure to give the right answer as indicating that relativistic mass has corrupted their brains.

My reaction on reading the question is that there's no reason to expect students to be able to construct an elementary argument if all they know is minimal SR and not the e.p. Am I missing some such elementary argument?

If you know the e.p., then I suppose you could reason that both A and B are being accelerated upward by a force from underneath their feet. Since they see each other as being at rest, and their displacement from each other is horizontal, their upward accelerations are equal, and must be caused by an equal force. This is all a little tricky because it does depend on the geometry. If A and B were displaced vertically from one another rather than horizontally, then I think the answer would be different. (This would be similar to the Bell spaceship paradox.)

Another approach would be to transform the earth's stress-energy tensor into A's frame -- obviously these students are not being expected to do that.

Am I missing some elementary argument that provides the correct answer? It seems to me that Oas's question is one that students at this level should not know the answer to, and that it's completely bogus to blame the result on relativistic mass.

2. Dec 29, 2013

### Ben Niehoff

The question is not what Bob measures Alice's mass to be, but what Bob observes Alice's scale to read. Obviously Alice's scale can only read one result at a time (and if Alice's mass is somehow changed, then the scale is also changed such that the reading is the same).

I don't think the equivalence principle is necessary. You just need the concept of rest mass, and the notion that Alice's rest frame is equally valid as Bob's.

"Relativistic mass" certainly does confuse things here, and ought to be thrown out in favor of "relativistic momentum", which is a vector quantity and makes more clear the fact that a scale oriented orthogonal to the direction of motion ought to give the same reading.

3. Dec 29, 2013

### Staff: Mentor

It's not just their displacement that has to be horizontal: their relative motion also has to be horizontal (i.e., perpendicular to the direction of acceleration). This isn't mentioned anywhere in the paper, but IMO it's a surprising omission, because including it would make the author's case much stronger. After all, if you're going to enumerate the problems with relativistic mass as a concept, surely one of the biggest is that it's not isotropic: the "transverse mass" (force applied perpendicular to the direction of relative motion) is *different* from the "longitudinal mass" (force applied parallel to the direction of motion).

(More precisely, the ratio of force to acceleration is different depending on the direction of the force vs. the direction of relative motion. The paper actually seems to define "relativistic mass" as the ratio of momentum to velocity; but that just means that a force parallel to the direction of relative motion breaks $F = ma$ instead of $p = mv$. Either way there's a failure of isotropy.)

I don't see any argument simpler than the one based on the EP that you give. I suppose that even if one didn't know the EP, one could reason that horizontal motion shouldn't affect a vertical force; but I think you would still have to have some physics background to be at all confident of such reasoning.

I also think it's interesting that the paper gives (=, =) as the "completely clorrect" answer without any discussion of the fact that, in GR, there are velocity-dependent correction terms to the Newtonian force, such that the reading on Al's scale actually *would* be larger than the reading on Bob's scale. Of course the corrections are extremely small for motion on the surface of the Earth, but they're there.

4. Dec 29, 2013

### Staff: Mentor

I think so... Both twins are looking at Al's scale so they're looking at the same needle lined up against the same number, or if it's a digital scale they're looking at the same LEDs illuminating.

5. Dec 29, 2013

### Staff: Mentor

I am not aware of an elementary argument. I would look at the proper acceleration for a stationary Rindler observer and a Rindler observer moving at constant velocity in one of the "non-accelerating" directions. I don't think that it is trivial.

6. Dec 29, 2013

### pervect

Staff Emeritus
I'd agree the Rinder problem is not trivial - from the work I did on the problem a while ago, I'd also have to disagree with the idea that both twins saw each other as being horizontally displaced. I'd also have to disagree about the scale readings, if I'm intepreting the argument correctly.

To recap the problem - we have an accelerating "spaceship observer", with an infinite flat floor in the spaceship frame, and another observer, a "sliding observer", sliding along the floor with a constant linear momentum as seen by the spaceship observer. The constant linear momentum condition winds up to be a constant proper velocity along the floor.

In particular posts #9 and #25. #25 is the most involved, but it actually comes up with a coordinate system with metric for the observer sliding along the elevator floor. It's got many counter-intuitive features, one of which is that a coordinate system attached to said observer that always points in the local "up" direction is rotating - this is seen by the presence of terms in the metric of the form

$g_{bb} \left(X\,dZ - Z \, dX\right)\,dT$

where T is the proper time of the sliding observer and Z always points "up", i.e. in the direction of local acceleration.

It can also be seen to be rotating by the fact that the Z vector is not fermi-walker transported.

One could use the coordinates I constructed to find Fermi Normal coordinates, but it turns out to be more convenient to use the ones I derived, I think - because in my opinion it's easier to work with metric coefficients that are not function of T as this respects the static nature of the problem. Transforming the rotation away is possible, but only at the expense of loosing the T-independence of the metric coefficients.

The shape of the spaceship floor is one of many things that could use more work. However, as I stated in the thread I think that the "flat" spaceship floor for the spaceship observer appears non-flat and rotating to the sliding observer. I expect that it should be an equipotential surface of the metric (a constant value of g_00). This can be argued by considering the fact that there is no redshift or blueshift between observers on the "flat" floor in the spaceship frame, so there can't be any redshift or blueshift in any other frame.

Alas, I can't guarantee that I haven't made an errors, and I suspect that the problem is difficult enough that nobody else has really confirmed all the calculations I made (or found all the typos).

7. Dec 29, 2013

### Staff: Mentor

I remember coming across this thread but not having time to do any calculations; but my thought at the time was to work the problem in standard Rindler coordinates, where the metric, assuming acceleration in the $Z$ direction, is (note that I've normalized so that the acceleration coefficient $g$ in the metric is 1--it would drop out of the calculation anyway so it's not important for this problem):

$$ds^2 = - Z^2 dT^2 + dX^2 + dY^2 + dZ^2$$

The only nonzero connection coefficients are $\Gamma^Z{}_{TT} = Z$ and $\Gamma^T{}_{TZ} = 1 / Z$, so computations "the old-fashioned way" are pretty simple. All we really need is the 4-velocity of the "sliding" observer, but that's just a boost in the $X$ direction:

$$u^a = \frac{\gamma}{Z} \partial_T + \gamma v \partial_X$$

where $v$ is the relative velocity and $\gamma = 1 / \sqrt{1 - v^2}$. (The factor of $1 / Z$ in the $\partial_T$ term is to ensure that the norm of $u^a$ is $-1$.) Then we just compute the proper acceleration for this 4-velocity:

$$a^a = u^b \nabla_b u^a = u^b \partial_b u^a + u^b \Gamma^a_{bc} u^c$$

Expanding out the nonzero terms (note that $v$ and $\gamma$ are constants so there are no nonzero partial derivatives, the only nonzero term comes from the connection coefficients):

$$a^Z = u^T \Gamma^Z{}_{TT} u^T = \frac{\gamma^2}{Z}$$

This is larger by a factor $\gamma^2$ than the corresponding term for the Rindler observer who is at rest in this chart. If this is correct, then the "completely correct" answer assumed in the paper referenced in the OP is actually wrong.

Last edited: Dec 29, 2013
8. Dec 29, 2013

### Staff: Mentor

This is also straightforward to verify using standard Rindler coordinates; we just need to compute the Fermi Derivative $D_F e^b = u^a \nabla_a e^b - \eta_{cd} e^c a^d u^b + \eta_{ef} e^e u^f a^b$, where $e^b$ is a unit vector in the $Z$ direction and $u^a$ and $a^a$ are the 4-velocity and the proper acceleration of the "sliding" observer from my previous post. This gives

$$D_F e^b = u^T \Gamma^T{}_{TZ} e^Z \partial_T - e^Z a^Z \left( \frac{\gamma}{Z} \partial_T + \gamma v \partial_X \right) = \frac{\gamma^3 v}{Z} \left( \frac{v}{Z} \partial_T - \partial_X \right)$$

This is clearly not zero (although I'm not sure exactly how it matches up with your results), so a vector in the $Z$ direction (which, as my previous post showed, is the *direction* of proper acceleration for the "sliding" observer, even though the magnitude is different than for the standard Rindler observer) is not Fermi-Walker transported along the "sliding" observer's worldline.

9. Dec 30, 2013

### pervect

Staff Emeritus
THis may or may not be a tangent, but I see the underlying issue here as "what is the best way to introduce students to relativistic dynamics".

Unfortunately, I don't have a good answer. The usual approach seems to be to ignore the issue totally. This results in students inventing their own (incorrect) notions of relativistic dynamics, for instance using Newtonian dynamics and replacing mass in the Newtonian equations with "relativistic mass'.

While this seems bad, I'm not sure what alternatives there are. Frankly, the only approaches I could recommend are rather abstract, that would be either the 4-vector approach, or the Lagrangian approach.

I think that these are the easiest routes I know of to correctly understanding relativistic dynamics, but both of them require a bit more sophistication that the introductory student to relativity will have. This suggests that the current approach of ignoring relativistic dynamics may be the best approach. Actually a slightly better approach would be to direct the student to books which do properly treat relativistic dynamics rather than totally ignore the issue.

Specific recommendatons here (which may not be the best).

Taylor-Wheeler "Space time Physics" for the 4-vector approach, and Goldstein's "Classical Mechanics" for the Lagrangian approach.

10. Dec 30, 2013

### phyti

Why can't it be a simple case of the sudents expecting an increase of mass due to speed resulting in a larger reading on the scale? This assumes an elementary and out of date concept, where additional energy invested gives diminishing returns, and was interpreted as inertial resistance.

11. Dec 31, 2013

### pervect

Staff Emeritus
Peter and I appear to be getting similar results - but I've taken the opportunity to add a bit of explanatory material at the end of https://www.physicsforums.com/showthread.php?t=701257 in case anyone is interested in slogging through the details. The main advantage of slogging through the details is that one winds up with an actual metric for the sliding observer, which can serve as a springboard to understanding the dynamics in the chosen coordinates.

However, it may not be worth the extra effort if one is satisfied with considering things from the perspective of the Rindler observer only. By calculating the right invariants, most questions can be answered from the more familiar Rindler perspective.