Does the weight of a relativistic mass change when measured on a moving scale?

[PAllen]

I find Moller's notation a bit confusing - from 12) on pg 290, second line, it appears that $\gamma_{lk}$ is the spatial part of $g_{lk}$, i.e. it's a 3x3 matrix, and that $u^k$ are the components of the 3- velocity, and $u$ is the magnitude of the velocity.

But then I don't see how to interpret $\gamma_k u^k$.

In any event ,if we assume that $\chi = 0$ we see that his relativistic mass is just m / sqrt(1 - v^2/c^2). Note that I implicitly assumed that the height was zero in my analysis, i.e. that $\chi=0$, I will now mention explicitly that that assumption is necessary for the analysis I did. I'm not quite sure what to make of the factor of two in the $\chi$ dependence offhand, not having seriously thought about any case other than $\chi=0$.

Much of what is needed here is defined in secion 92, on page 245. I have not yet studied this enough for my own conclusions, but did find that section 92 is where he first clearly defines these notations.

 

[PeterDonis]

if we assume that $\chi = 0$ we see that his relativistic mass is just m / sqrt(1 - v^2/c^2). Note that I implicitly assumed that the height was zero in my analysis, i.e. that $\chi=0$,

I'm not sure Moller's $\chi$ is height. I think it's the Newtonian "gravitational potential", i.e., for the static, spherically symmetric case, with the standard normalization (which I don't think Moller is using--see below), it would be $\chi = - GM / r$. MTW uses $\phi$ to denote the same thing. The factor of 2 gives the equivalence $1 + 2 \chi / c^2 = 1 - 2 GM / c^2 r$, which of course is familiar.

Having read through Moller's analysis in more detail, I think what he is trying to capture with his "relativistic mass" in a gravitational field actually is somewhat related to energy at infinity (despite what I said before), but with a weird normalization of it. For the case $v = 0$ (i.e., a static object), his equation reduces to $m = m_0 / \sqrt{1 + 2 \chi / c^2}$. If we use the formula I gave above for $\chi$ we get $m = m_0 / \sqrt{1 - 2 GM / c^2 r}$. As I pointed out in an earlier post, this is backwards from the way we ordinarily think about energy at infinity, because we expect it to be smaller than the rest mass for an object that's static at a finite $r$.

However, if we suppose that $\chi = 0$ corresponds, not to infinity, but to some finite height (such as the surface of the Earth), and that $m_0$ is the locally measured invariant mass of an object at rest at that height, then $m$ (when $v = 0$) is just the locally measured invariant mass of the same object at a height corresponding to the potential $\chi$, normalized as I just described. In other words, $m - m_0$ is just the work done to raise the object from $\chi = 0$ to $\chi$.

This all makes sense, although it seems to me to be a strange use of the term "relativistic mass"--but of course I would not choose to use that term at all.

 

[nitsuj]

What you should read about is invariant mass.

ahaha, Thank you PAllen, as soon as the wiki page loads, I see "Center of momentum" highlighted as a link. that simple term makes allot of sense in this "rest mass" context, in another discussion center of momentum concept popped up in a comparative motion context.

Anyways in the example of an Atom, and what makes up it's mass, or more specific to the OP a "weighty" feel you can hold in you hands or on a scale "Kinetic and Gluon Field Energy" apparently is the majority of an atoms mass.

 

[pervect]

It looks like $\chi$ is defined on pg 246 in 94, and is referred to as a "dynamical gravitational potential. It appears that as long as $g_{44}$ has a unit magnitude, $\chi$ is equal to zero. Moeller mentions explicitly that the value of $\Gamma$ is a generalization of the Lorentz time dilation factor, so we expect it to reduce to the later if we adopt appropriate simplifications. The most obvious simplification is to assume $\chi=0$, which is equivalent to assuming $g_44$ is -1, in which case we see by inspection that the expression is equal to $\Gamma$.

Looking more closely at the derivation, though, Moeller's starting point for the equations in question appears to be that the covariant derivative of the momentum 4-vector is equal to zero for a particle in free fall, see the discussion of (9) on 289.

where $\frac {D \, P^i}{d\tau} =$ ... represent the covariant derivatives of the four-momentum vector

Unfortunately, what we're really interested in is the force on a particle that's not in free fall. It's fairly well known that the 4-force is the covariant derivative (with respect to proper time) of the momentum for a particle not in free fall, though you'd need some other section of the text to actualy explicitly justify this.

Covariant derivaties are overkill for this problem, we can eliminate the need for covariant differentiation by working in an inertial frame. At the most general level, then, we are left with the idea that in an inertial frame force is rate of change of momentum with respect to time.

There are a couple of different specific ways to desribe forces. One way uses the formalism of 4-vectors, and to measure time using proper time. This has the strong advantage that conceptually the 4-force is the same regardless of one's coordinate choice, though the components may vary. This happens because 4-vectors are observer independent objects and proper time is a coordinate independent scalar. The observer independence of the formalism is a HUGE advantage, one that cannot in my opinion be overstated.

The other way that may be seemingly more familiar can be more confusing, because one has to be careful to specify the frame of reference one is using to measure the force and momentum. This scheme says that the 3-force is the derivative of the 3-momentum with respect to the observer dependent coordinate time. (The process of going from 4-momentum to 3-momentum is very simple one drops the extra component. The other difference is using proper time rather than coordinate time.)

However, given that a scale measures the familiar 3-force, if one wants to explain the results in terms of scale readings, I see no alternative but to take into account the coordinate dependence even though it's confusing.

 

[pervect]

A rather important correction - the second one :(. I finally looked up the old thread where I worked out the proper acceleration of the sliding block. The thread was https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/

The proper acceleration of the sliding block is then proportional to $\gamma_0^2$, not $\gamma$, my memory was inaccurate on this part. $\gamma_0$ is given by the usual formula, 1 / sqrt( 1 - v^2/ c^2), where v can be thought of as the constant sliding velocity of the sliding block as measured by an observer on a frame that is instantaneously co-moving with the elevator floor.

The inertial coordinate system used in the problem analysis of the above post is an inertial frame where the elevator accelerates in the z direction, and the block slides in the x direction. In that frame, the z-velocity of the elevator floor is assumed to be zero at t=0, and z=0 is the location of the elevator floor at t=0.

In said inertial coordinate system, $\gamma_0$ is the x-velocity of the block in the specified inertial frame at t=0. The time needs to be specified because the x- velocity of the block is not constant in time in the global inertial coordinate system used, only the value of the x-velocity in a frame instantaneously co-moving with the elevator floor is constant.

The rest of my remarks still apply, the ratio of the force as measured by an observer on the sliding block to the force as measured by means of a "weight in motion" system based on the floor pressure is still $\gamma_0$. However, because of the previous error, the scale reading on the sliding block will be $\gamma_0^2 \, m \, g$ for a scale mounted on the block, and the scale reading of a scale mounted on the floor will be $\gamma_0 \, m \, g$. g being the proper acceleration of the elevator.