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

[PAllen]

Well let's see ... Let gravity try to deflect a fast moving electron, whose path is kept straight by suitable electric field, electric field strength is proportional to the weight of the electron.

Or: Measure the weight of that device when an electron is in, subtract the weight of an empty device, result is the weight of the electron.

This looks like putting a fast electron on a scale to me.

I agree that is an example of adding something else, to complete the description. Pervect also added other ingredients to complete the description (leading to two different answers depending on how he did it). Most of the initial responses effectively changed the problem to something different than a literal reading of the OP, because it had not been suggested how to formalize the what the OP was getting at. We still have the conclusion that variations of the problem that are equivalent at low relative speeds differ at high relative speeds. In particular, using deflection as a measure, we still have that there will be an extra factor besides gamma (approaching 2).

 

[Jonathan Scott]

So let's put mirrors above the north and south poles of a black hole, at the photon sphere. A light pulse bounces between the mirrors pushing the mirrors. The weight of the light is the force exerted on the mirrors.

Now we move the mirror at the south pole to the equator. This operation makes the force on the north pole mirror alone same as the total force on both mirrors in the original configuration. The weight of light increased.

If you're assuming that the mirrors in this case are locally vertical, reflecting the photon back around the photon sphere, that doesn't match this case, as the light is being allowed to fall freely and is effectively in orbit. For this model, the mirrors have to be angled in such a way that a "horizontal" line between them is curved by an amount matching the curvature of space over that distance (which is only about half of the contribution to the curvature of the photon sphere, depending on the coordinate system). We have also been assuming it's small enough that the direction of the source is effectively the same from both mirrors, otherwise there are other effects as well.

If you're saying that the same amount of energy bouncing back and forwards over a shorter distance bounces more frequently and hence imparts more force, then that's correct for the lateral force on the mirrors. However, for the mirrors to match the local vertical at the mid-point of the travel taking into account curvature of space, then for the shorter trip the angle between the mirrors is smaller, so the vertical component of the rate of change of the momentum is still the same for the same amount of energy.

 

[PAllen]

An amusing example of how intuitions can lead astray, in pure SR, is that suppose you set up the constraints of the 'particle sliding on a scale' as follows:

1) The floor is undergoing Born rigid, uniform acceleration orthogonal to its surface. This means, it is describe by 'y' slices of given t, in Rindler coordinates.
2) In a starting inertial frame, the y speed of the sliding particle remains constant (otherwise, one reasons, it is undergoing acceleration parallel to the floor, which is at odds with sliding without friction).

This description is leads to a surprising conclusion. Starting from the floor momentarily at rest in some initial inertial frame, the constraints can no longer be met after time of H/vγ (in units with c=1, where H=1/acceleration, i.e. the distance between the Rindler Horizon and the floor). At that time, the particle would be required to be moving with a light like tangent. After that t, these constraints amount to a spacelike trajectory. Further, weight measured by the scale approaches infinity as this critical time is approached.

Thus, the particle sliding on a scale as modeled by Pervect (which uses - very reasonable - local constraints, rather the the global ones suggested by simple intuition) entails a required 'relativistic drag' force, parallel to the floor, as viewed from some starting inertial frame. One description of Pervect's set up is to moment to moment restate the problem relative to an inertial frame in which the floor is at rest. This is very reasonable, because it accepts that in SR, the definition of something like sliding with no friction is best stated with a local definition.

 

[nitsuj]

What is the source of rest mass? What is the source of mass of protons/neutrons?

Wiki says "The remainder of the proton mass is due to the kinetic energy of the quarks and to the energy of the gluon fields that bind the quarks together."

the quarks are about 1% of the mass of the proton. No kidding relativistic mass is confusing with respect to "rest" mass. Is that kinetic energy & gluon field seen as "rest mass" able to be describe by SR as relativistic mass?

 

[PAllen]

What is the source of rest mass? What is the source of mass of protons/neutrons?

Wiki says "The remainder of the proton mass is due to the kinetic energy of the quarks and to the energy of the gluon fields that bind the quarks together."

the quarks are about 1% of the mass of the proton. No kidding relativistic mass is confusing with respect to "rest" mass. Is that kinetic energy & gluon field seen as "rest mass" able to be describe by SR as relativistic mass?

What you should read about is invariant mass. Two examples (described in some given frame):

1) Two identical particles are moving parallel at the same speed. The invariant mass is the sum of their rest mass.

2) Two identical particles are moving toward each other at the same speed. The invariant mass is the sum of their KE / c² + the sum of their rest mass.

Invariant mass reflects the fact that in the latter case, the kinetic energy is available (via collision) to be be converted to rest mass of bound state, while in the former case it is not. However, the sum of 'relativistic' mass in both cases is the same. To me, this is an example of why invariant mass is very useful, while relativistic mass is not.

Invariant mass, mathematically, is the norm (magintude) of total 4-momentum. 4-momentum has the virtue it simply adds as a vector. If you want to know the 'system rest mass' of bunch of particles, just add their 4-momenta and compute the norm.

 

[harrylin]

Consider two equal masses, m, are placed on two identical scales in a uniform gravitational field. If one of the masses moves with relativistic speed, does the moving scale read mg or γmg? Why?

It seams that people used to call γm the relativistic mass where m is the rest mass. However, more recent authors think this is a faulty terminology since there is no such thing as relativistic mass, only relativistic momentum. I have asked my professor and he said the weight does not change, but I am not satisfied and looking for a reason why that might be so.

I'm a bit slow, and so I now found a GR formula by Moller (The Theory of Relativity, p.191) which somewhat relates to your first question; you may find it interesting.

He finds for the "important case" that the coordinate system is "time orthogonal" (sorry I'm not sure what that means!), without approximation:

m = m₀ / sqrt (1 + 2X/c² - v²/c²)

in which
m = relativistic mass
X = gravitational potential

That's a neat equation. :)
And to exclude any possible misunderstanding:

K = m G = -m grad X
i.e. the gravitational force is connected with the scalar gravitational potential in the same way as in Newton's theory.

 

[PAllen]

I'm a bit slow, and so I now found a GR formula by Moller (The Theory of Relativity, p.191) which somewhat relates to your first question; you may find it interesting.

He finds for the "important case" that the coordinate system is "time orthogonal" (sorry I'm not sure what that means!), without approximation:

m = m₀ / sqrt (1 + 2X/c² - v²/c²)

in which
m = relativistic mass
X = gravitational potential

That's a neat equation. :)
And to exclude any possible misunderstanding:

K = m G = -m grad X
i.e. the gravitational force is connected with the scalar gravitational potential in the same way as in Newton's theory.

It looks interesting but I can't find it at:

https://archive.org/stream/theoryofrelativi029229mbp#page/n207/mode/2up
Since there are presumably many editions of this book with different pagination,
I would appreciate greatly if you were able to locate the corresponding section in this online source, as I would like to read the context.

 

[harrylin]

It looks interesting but I can't find it at:

https://archive.org/stream/theoryofrelativi029229mbp#page/n207/mode/2up
Since there are presumably many editions of this book with different pagination,
I would appreciate greatly if you were able to locate the corresponding section in this online source, as I would like to read the context.

Thanks for the online reference! In my library book it's in chapter X, on Physical Phenomena, §110. Oh now I see, typo on page number (sorry): p.291.

PS there is another subtlety that I overlooked: he writes u instead of v, and it's not clear if that makes a difference for this case.

 

[PeterDonis]

He finds for the "important case" that the coordinate system is "time orthogonal" (sorry I'm not sure what that means!), without approximation:

m = m0 / sqrt (1 + 2X/c2 - v2/c2)

in which
m = relativistic mass
X = gravitational potential

I think that by "time orthogonal" he means that the spacetime is static and the the time coordinate is such that each spacelike hypersurface of constant time is orthogonal to every integral curve of the time coordinate (which can only happen if the spacetime is static, so it has a hypersurface orthogonal timelike Killing vector field).

If the above is correct, then I think you may have miscopied the formula. I think the intent of the formula is to give what is usually called "energy at infinity" for a test object moving on a geodesic in a static gravitational field (because it looks like it's supposed to be the "time" constant of the motion for the test object, i.e., the "time" component of the object's 4-momentum, which is a constant for geodesic motion because of the timelike Killing vector field). If $m$ is supposed to be energy at infinity, and $m_0$ is rest mass, then the RHS should not be a ratio, it should be a product, since the energy at infinity of a bound object is smaller than the object's rest mass. However, it's possible that I'm misinterpreting what the formula is supposed to represent.

 

[harrylin]

I think that by "time orthogonal" he means that the spacetime is static and the the time coordinate is such that each spacelike hypersurface of constant time is orthogonal to every integral curve of the time coordinate (which can only happen if the spacetime is static, so it has a hypersurface orthogonal timelike Killing vector field).

If the above is correct, then I think you may have miscopied the formula. I think the intent of the formula is to give what is usually called "energy at infinity" for a test object moving on a geodesic in a static gravitational field (because it looks like it's supposed to be the "time" constant of the motion for the test object, i.e., the "time" component of the object's 4-momentum, which is a constant for geodesic motion because of the timelike Killing vector field). If $m$ is supposed to be energy at infinity, and $m_0$ is rest mass, then the RHS should not be a ratio, it should be a product, since the energy at infinity of a bound object is smaller than the object's rest mass. However, it's possible that I'm misinterpreting what the formula is supposed to represent.

I did make a typo, he has u instead of v, while he has v in the SR section. It will be interesting to be sure how it should be interpreted. For sure the space-time is static indeed. Is it about a particle falling straight down? Here's the page: https://archive.org/stream/theoryofrelativi029229mbp#page/n307/mode/2up

 

[PeterDonis]

Here's the page

Ok, I was mistaken, he's not trying to describe energy at infinity. I'm not exactly sure how what he calls "relativistic mass" would be described in modern terms; I'll have to think about this some more.

 

[PAllen]

Ok, I was mistaken, he's not trying to describe energy at infinity. I'm not exactly sure how what he calls "relativistic mass" would be described in modern terms; I'll have to think about this some more.

I started looking through this, and it seemed, to really understand it his notations and concepts in context, you should start from section 92, p. 245, where he introduces his notions of dynamic gravitational potentials. Some of how he divides the metric into different pieces, especially his notion of spatial metric at a point, seems equivalent to Landau & Lifschitz.

 

[jartsa]

If you're assuming that the mirrors in this case are locally vertical, reflecting the photon back around the photon sphere, that doesn't match this case, as the light is being allowed to fall freely and is effectively in orbit. For this model, the mirrors have to be angled in such a way that a "horizontal" line between them is curved by an amount matching the curvature of space over that distance (which is only about half of the contribution to the curvature of the photon sphere, depending on the coordinate system). We have also been assuming it's small enough that the direction of the source is effectively the same from both mirrors, otherwise there are other effects as well.

If you're saying that the same amount of energy bouncing back and forwards over a shorter distance bounces more frequently and hence imparts more force, then that's correct for the lateral force on the mirrors. However, for the mirrors to match the local vertical at the mid-point of the travel taking into account curvature of space, then for the shorter trip the angle between the mirrors is smaller, so the vertical component of the rate of change of the momentum is still the same for the same amount of energy.

Oh dear I forgot that it's quite normal that long things have "reduced weight" in non-uniform gravity fields. (Like rigid circle around a planet is "weightless")

And space is not as curved as the path of light, like I thought? Rigid rods connecting two parallel mirrors don't bend as much in a gravity field as light beam bends?

 

[Jonathan Scott]

Oh dear I forgot that it's quite normal that long things have "reduced weight" in non-uniform gravity fields. (Like rigid circle around a planet is "weightless")

And space is not as curved as the path of light, like I thought? Rigid rods connecting two parallel mirrors don't bend as much in a gravity field as light beam bends?

Yes, that sounds correct to me.

 

[pervect]

I'm a bit slow, and so I now found a GR formula by Moller (The Theory of Relativity, p.191) which somewhat relates to your first question; you may find it interesting.

He finds for the "important case" that the coordinate system is "time orthogonal" (sorry I'm not sure what that means!), without approximation:

m = m₀ / sqrt (1 + 2X/c² - v²/c²)

in which
m = relativistic mass
X = gravitational potential

That's a neat equation. :)

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 $\gamma_k u^k$.

[add]I did find this eventually, it's defined in (63) on pg 238. A few editorial comments, Moeller is an old textbook, and uses relativistic mass as many old textbooks do. I'm not a fan of relativistic mass, and I find the old textbook treatments that use it like Moeller's clunky and hard to follow. As long as the treatments are correct, <snip irrelevant stuff> I don't have a huge argument against relativistic mass. I still don't like it much, and find it hard to follow, a drain on my time, and something that I might not follow very closely, but the important part is getting the right answer.

Going back to the paper - ,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 $\chi=0$.

 

[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.