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

[MiladP]

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.

 

[Orodruin]

You are trying to treat Newtonian gravitational effects with special relativity. This will not work out well.

By construction, special relativity does not include gravitational effects. If you go to general relativity that does treat gravity, it is not mass, but energy and momentum, that is the source for gravity. Yet, the question you are posing will depend on what you are actually meaning when you say "put it on a scale".

Also see our https://www.physicsforums.com/posts/4919337/ .

 

[harrylin]

Welcome on Physicsforums! :)

Here are my 2cts.
Reformulating your question to an inertial reference system in which the scale is in rest, you are in fact asking about the effect of moving the source of gravitation at relativistic speed. That is, as you may guess, a question that cannot be answered by special relativity: it is part of general relativity. I doubt that you wanted to ask a GR question.
Here's how you may change your question to make it better fitting for the title that you gave this topic, and also part of special relativity:

Consider a mass, with rest weight m, placed on a scale in a uniform gravitational field such as on a table on Earth, with a weightless circular container around it (for simplification). The scale rests in place on the table. If the mass is now made to move with relativistic speed in a circle on top of the scale, does the scale read mg or γmg? Why?
Note: in view of balance issues you could represent the mass as a ring-shaped object that is made to spin; then you can also can get rid of the weightless container. :)

Likely you can figure out the correct answer.

 

[Orodruin]

Here's how you may change your question to make it better fitting for the title that you gave this topic, and also part of special relativity:

Consider a mass, with rest weight m, placed on a scale in a uniform gravitational field such as on a table on Earth,

The highlighted parts are contradictory, there are no gravitational fields in special relativity.

 

[harrylin]

The highlighted parts are contradictory, there are no gravitational fields in special relativity.

Apparently there are different definitions of SR; I mean with it the relativity theory of 1905 and that theory does give an answer. It would be interesting if GR gives a different answer!

 

[Orodruin]

Feel free to provide a reference to where Einstein mentions gravity in his original 1905 paper.

 

[Nugatory]

Apparently there are different definitions of SR; I mean with it the relativity theory of 1905 and that theory does give an answer. It would be interesting if GR gives a different answer!

GR gives the same answer as your procedure (in the weak field approximation).

It is legitimate to treat gravity as an ordinary Newtonian force and apply the methods of SR as long as you use the weak field approximation, the observed velocities are small compared to that of light, and the setup is local enough that tidal effects can be ignored. I believe that all of these elements are present in your ring-in-a-box hypothetical.

 

[DrStupid]

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?

According to https://home.comcast.net/~peter.m.brown/ref/mass_articles/Olson_Guarino_1985.pdf it will read

\left( {1 + \beta ^2 } \right) \cdot \gamma \cdot m \cdot g

if the velocity is almost constant. For a circular motion (as suggested by harrylin) you will most probably get another result.

 

[PAllen]

According to https://home.comcast.net/~peter.m.brown/ref/mass_articles/Olson_Guarino_1985.pdf it will read

\left( {1 + \beta ^2 } \right) \cdot \gamma \cdot m \cdot g

if the velocity is almost constant. For a circular motion (as suggested by harrylin) you will most probably get another result.

Well, not quite. Scales are not mentioned in that paper, and it is hard to imagine how you put a flyby object on a scale. What that paper shows is that if you compare:

1) The prediction of Newton's law of gravitation applied to the deflection on a small test body by a flyby object, using the flyby object's rest mass as the mass in Newton's formula

with:

2) The deflection predicted by GR

the latter is larger by (1+β²)γ

Harrylin's proposal, which could actually be done with a scale, would instead show an increase by γ, suggesting energy as a source of gravity.

On the other hand, if you have two rapid flyby objects going in the same direction, and you observe their mutual deflection, you would find that their mutual deflection approaches zero as their speed approaches c.

I claim this all goes to support the position that these questions are ill formed, and that trying to treat relativistic mass a source of gravity is doomed to provide confusion more than clarity.

 

[pervect]

I don't have a detailed reference, but it's not too hard to calculate that if you attempt to realize a "uniform gravitational field" by means of an elevator, the elevator floor will be flat in the frame of an inertial observer, and also flat in the frame of the observer who is standing still on the elevator, but the floor will not be flat in the frame of an observer sliding along the floor. This is due to the relativity of simultaneity - the accelerating trajectory is not linear in time, and when you apply the usual Lorentz boost to boost to the appropriate instantaneously co-moving inertial frame, the result is a non-flat surface.

By "frame" here I mean the frame of an instantaneously co-moving inertial observer.

I've got an old thread on the topic with some detailed calculations, (I suppose in some sense it's a detailed reference, except that it's not a textbook reference). Wanabee Newton (I think) had some remarks about textbook exercises which came up with the same observation.

Anyway, the floor not being flat helps reconcile all the scale readings, much as you expect to weigh more when you're on a roller coaster that's moving along a curved track at a high velocity.

There's also an interesting effect that causes rotation of the sliding observer in this case, i.e. his basis vectors are not Fermi-walker transported.

 

[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? [..] 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.

[..] Here's how you may change your question to make it better fitting for the title that you gave this topic [..]:
Consider a mass, with rest weight m, placed on a scale in a uniform gravitational field such as on a table on Earth, with a weightless circular container around it (for simplification). The scale rests in place on the table. If the mass is now made to move with relativistic speed in a circle on top of the scale, does the scale read mg or γmg? Why? [..] Likely you can figure out the correct answer.

OK then, here's also my analysis of the rephrased question, using only special relativity. I think that this does not require the use of any coordinates or mathematics.

The mass that is discussed in the theory happens to be defined by means of gravity, and at the time it was assumed that atoms are composed of moving particles. This implies that according to SR - just as in classical mechanics - if one determines the mass of the container in rest, by means of its inertia, then this should agree with the same as determined by means of its weight.

 

[DrStupid]

What that paper shows is that if you compare:

1) The prediction of Newton's law of gravitation applied to the deflection on a small test body by a flyby object, using the flyby object's rest mass as the mass in Newton's formula

with:

2) The deflection predicted by GR

the latter is larger by (1+β²)γ

And that's what the scale would measure if it prevents the mass from deflection.

Harrylin's proposal, which could actually be done with a scale, would instead show an increase by γ, suggesting energy as a source of gravity.

The source of gravity is not energy but the stress-energy-tensor. The increase by γ is limited to systems without momentum or angular momentum.

 

[PAllen]

And that's what the scale would measure if it prevents the mass from deflection.
The source of gravity is not energy but the stress-energy-tensor. The increase by γ is limited to systems without momentum or angular momentum.

1) Probably, but that is not what the paper derives, and does not constitute weighing the flyby object on a scale.

2) I'm fully aware of that, and that was the gist of my post, showing different cases, only one of which was just γ. Perhaps, I was wrong about Harrylin's case, but a similar one that is well known is a box of inelastically bouncing, identical, particles with given speed. The effective inertial and gravitational mass of the box (assuming the box itself has negligible mass) is given γM, with M being the sum of the rest masses of the particles. I guessed Harrylin's case would be similar, but you are right the angular momentum might change it noticeably [the net momentum in any given volume element is not zero, as per the gas case], and one should really set up an SET for it (given that the system is stationary, it should be ok to compute Komar mass rather than having to deal with ADM or Bondi energy).

 

[jartsa]

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?

That's easy to calculate. We use special relativity to calculate what the scales read in an accelerating rocket, the answer is γmg. Then we use the equivalence principle, and note that the answer is the same in a homogeneous gravity field.

When a myon passes the sun, the weight of the myon seems to be more than γmg, but that's because the gravity field causing the weight is not homogeneous.

 

[pervect]

I claim this all goes to support the position that these questions are ill formed, and that trying to treat relativistic mass a source of gravity is doomed to provide confusion more than clarity.

Treating relativistic mass as the source of gravity is definitely wrong, and it is probably in many cases the motivation for asking the questions :(. Even given that issue, a detailed answer to the question after it is properly nailed down (which can be hard) can be useful to the careful reader. Probably these answers will not as much use as taking a course in GR, but then again the investment of time is considerably less. The biggest issue in practice is that frequently points that turn out to be key aren't asked at all, or that the reader has an intuitive idea of the answer which turns out to be incorrect when relativistic effects (particularly the relativity of simultaneity) are taken into account.

 

[PAllen]

That's easy to calculate. We use special relativity to calculate what the scales read in an accelerating rocket, the answer is γmg. Then we use the equivalence principle, and note that the answer is the same in a homogeneous gravity field.

When a myon passes the sun, the weight of the myon seems to be more than γmg, but that's because the gravity field causing the weight is not homogeneous.

Unfortunately, this argument establishes exactly nothing. A rocket would see a rapidly moving particle entering through a pin hole reach the other side of the rocket consistent with the rocket's experienced acceleration. However, as long as one assumes that gravitational mass and inertial mass are the same, this tells us exactly nothing about the mass of the particle. It could be 42m for all we know.

 

[jartsa]

Unfortunately, this argument establishes exactly nothing. A rocket would see a rapidly moving particle entering through a pin hole reach the other side of the rocket consistent with the rocket's experienced acceleration. However, as long as one assumes that gravitational mass and inertial mass are the same, this tells us exactly nothing about the mass of the particle. It could be 42m for all we know.

Now I don't understand. We do OP's experiment, but not in a uniform gravity field, but in an accelerating rocket. Then we deduce that in a small laboratory on the Earth the experiment goes the same way, because the gravity field is uniform enough in a laboratory small enough.

I don't know the details of the experiment, but a fast moving mass was put on a scale, according to the OP.... I also said something about irrelevant about extra weight of myons passing the sun. Let's just forget that part.

 

[PAllen]

Now I don't understand. We do OP's experiment, but not in a uniform gravity field, but in an accelerating rocket. Then we deduce that in a small laboratory on the Earth the experiment goes the same way, because the gravity field is uniform enough in a laboratory small enough.

I don't know the details of the experiment, but a fast moving mass was put on a scale, according to the OP.... I also said something about irrelevant about extra weight of myons passing the sun. Let's just forget that part.

You haven't put a moving weight on a scale. In fact, so far, no one has suggested a way that is possible. I don't see a way that it is. Various ways to get to effective gravitational mass of a rapidly moving object (other than putting it on a scale) were suggested, because of this impossibility. Different alternatives produce different answers.

My argument is simply that deflection in a rocket or 'near uniform' gravity per se, does not tell you anything about mass (because all masses will have same deflection in this setup). Maybe if you supplement this with other arguments you can get somewhere.

 

[PAllen]

I won't have time to play with this for a while, but I've noticed something interesting. I can't get from here yet to a principle of equivalence + SR argument heuristically justifying the γ(1+β²), but I suspect there may be one. I note that:

γ(u⊕u) = γ(u)²(1+β(u)²)

so some argument from different frames, with time dilation may 'justify' the paper's result with SR + POE.

 

[Jonathan Scott]

The factor $(1 + \beta^2)$ due to space curvature in GR should NOT appear in the weight because a horizontal plane in the scales will already follow the local curvature of space. It would appear in the motion described relative to a flat coordinate system, for example when computing an orbit.

 

[Jonathan Scott]

This even applies to a light beam; a light beam reflected between mirrors which appear to be parallel and vertical in a local frame is accelerated downwards with the same acceleration as a brick.

 

[PAllen]

The factor $(1 + \beta^2)$ due to space curvature in GR should NOT appear in the weight because a horizontal plane in the scales will already follow the local curvature of space. It would appear in the motion described relative to a flat coordinate system, for example when computing an orbit.

The paper cited in this thread (and in many threads throughout PF), gets this factor for a local flyby. A scale is not involved. Possibly, curvature still is because the interaction involves a path through a large part of the field of a moving mass.

Also, it is worth noting that this same factor appears in analyses of the attraction by a beam of relativistically moving dust on a test particle. It is possible that this is also attributable, in some less obvious way, to curvature.

 

[PAllen]

This even applies to a light beam; a light beam reflected between mirrors which appear to be parallel and vertical in a local frame is accelerated downwards with the same acceleration as a brick.

But a light beam's affect on a test particle does have a factor of 2.

 

[Jonathan Scott]

But a light beam's affect on a test particle does have a factor of 2.

Yes, that makes sense relative to a flat coordinate system. If you look at the light reflecting off locally vertical mirrors relative to a flat coordinate system, then the mirrors are slightly angled upwards because of curved space, effectively halving the downward rate of change of momentum (and being pushed down by the light in return). So the actual rate of change of momentum of the light caused by gravity relative to the flat coordinate system, including the effect of curvature of space, is twice the apparent rate of change in the local frame, so we would expect the equal and opposite effect in that coordinate system also to be equivalent to twice the local force.

 

[Khashishi]

Consider the case where the gravitational field is generated by a thin infinite plane of density rho (with normal pointed in the z direction). The stress energy tensor looks like
$\begin{pmatrix}\rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
A mass moving in the x direction sees a transformed plane in its rest frame:
$\begin{pmatrix} \gamma^2 \rho & -\gamma^2 \beta \rho & 0 & 0 \\ -\gamma^2 \beta \rho& \gamma^2 \beta^2 \rho & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
Assume the scale is moving in the x direction and the mass is sitting on top of it. It should be possible to calculate the proper acceleration on the mass in some weak field limit, and then the weight is the force needed to generate that acceleration is just $F=m_0a$. But I don't know how to do the calculation.

 

[PAllen]

Consider the case where the gravitational field is generated by a thin infinite plane of density rho (with normal pointed in the z direction). The stress energy tensor looks like
$\begin{pmatrix}\rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
A mass moving in the x direction sees a transformed plane in its rest frame:
$\begin{pmatrix} \gamma^2 \rho & -\gamma^2 \beta \rho & 0 & 0 \\ -\gamma^2 \beta \rho& \gamma^2 \beta^2 \rho & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
Assume the scale is moving in the x direction and the mass is sitting on top of it. It should be possible to calculate the proper acceleration on the mass in some weak field limit, and then the weight is the force needed to generate that acceleration is just $F=m_0a$. But I don't know how to do the calculation.

This kind of turns the OP on its head. The scale becomes the source of gravity, rather than being a passive instrument.

 

[pervect]

The factor $(1 + \beta^2)$ due to space curvature in GR should NOT appear in the weight because a horizontal plane in the scales will already follow the local curvature of space. It would appear in the motion described relative to a flat coordinate system, for example when computing an orbit.

I mostly agree, except that I would say that it's the space-time and not space that's flat in the case of the accelerating rocket, i.e. the Riemann tensor is zero. If you want to claim space is flat, you need to specify the time-slicing. Thus if you specify that you're using Rindler coordinates, you can say that the spatial part of the metric is flat with that time slicing.. PIcky, I know, but when you consider that the floor of the rocket can be curved to some observers and not-curved to others because of differences in the simultaneity conventions, it's probably needed.

As far as resolving the weight issue, I would suggest just calculating the 4-acceleration of the sliding observer which one can take to be pointlike then arguing that the "weight" that the observer would experience is just their 4-acceleration. I've done this myself , and I believe at least one other person has as well.

 

[MiladP]

Thank you everyone for your replies. Many of you disagree that this is a SR subject. I claim that it is legal to discuss gravity in a SR context as long as it is uniform and the object under question does not accelerate. However, I agree with Jonathan even though his answer involves GR (thanks). I will update my knowledge on GR and meanwhile I look for SR answers from you :) Please read my next post for some interesting scenarios related to this question.

 

[MiladP]

As a reminder, my goal is to evaluate the physical meaning of the following equation m = γ m₀. In other words, does mass actually change or we write this equation because we can express momentum as p = m v? What does it mean for the mass to change?

Here is a similar thought experiment that gets rid of gravity (but I am not sure if it addresses the above questions). Consider two equal masses attached to two identical spring far from any other object. One mass is moving with a relativistic velocity wrt the other one. The two springs are parallel and the direction of motion is perpendicular to the springs. (If you are wondering what are the springs attached to, just consider performing this experiment in a heavy spaceship). Now we oscillate both masses (no it is not GR because I'm not talking about the masses ). The period T of the two systems are related by T_{1 seen by 1} = γ T_{2 seen by 1}, which we can write as m₁/k₁ = m₂/(k₂ (1 + β²)). Since the springs are perpendicular to the direction of motion of the second system, k's don't change (not so sure about this), then m₂ = (1 + β²) m₁. Whaaaaaat?

 

[Orodruin]

As a reminder, my goal is to evaluate the physical meaning of the following equation m = γ m0.

I suggest you start by reading the FAQ I linked to in post #2.

 

[pervect]

Thank you everyone for your replies. Many of you disagree that this is a SR subject. I claim that it is legal to discuss gravity in a SR context as long as it is uniform and the object under question does not accelerate.

I'm not quite sure where you got this idea. You can apply SR in any situation where you have a flat space-time, regardless of whether the objects are accelerating or not.

It's unclear what you think you mean by "uniform gravitational field". If you call the pseudo-gravitational field seen by an observer riding Einstein's elevator "a uniform field", in spite of the fact that the acceleration of a test particle in said "uniform" field depends on its height then you CAN" apply SR to this situation. I more or less assumed that this was what you meant on the first read of the post, but now that I've read more it's unclear what you actually thought you meant when you said "uniform gravitational field".

So the way to ask your question in SR from my perspective is to ask "If you have an Einstein's elevator, whose floor accelerates upwards with some proper acceleration g, what is the proper acceleration of an object sliding across the elevator floor".

To compute the magnitude of the proper acceleration, it is convenient to understand 4-vectors in general and 4-accelerations in particular. But you might be able to manage if you understand at least what a proper acceleration is. I can't tell if you understand this concept or not - by "understand" I mean a shared understanding, so that we think of the same concepts when we read the same words.

Being a bit jaded, I'm rather suspecting you don't understand the terminology (or jargon, if you're unfamiliar with it). But not much communication is going to happen if we don't share the same language.

I also suspect from some remarks that you made that one of your goals is to understand relativistic dynamics. If that is the case, you're going about it with some preconceived notions that is going to make that task more difficult. Additionally, understanding dynamics in special relativity is a good first step, but it's not going to tell you anything about gravity, which may be another of your goals. To understand gravity, you will have to go beyond SR, and also beyond the idea that "mass" is what causes gravity.

 

[DrStupid]

As a reminder, my goal is to evaluate the physical meaning of the following equation m = γ m₀. In other words, does mass actually change

According to the current naming convention "mass" is rest mass and therefore frame independent.

or we write this equation because we can express momentum as p = m v?

Delete "can" and it makes sense. If you write momentum as p=m·v then you will get m=γ m₀. But as mentioned above "m" is not called "mass" and many physicist don't even talk about it. Better use E/c² instead.

 

[jartsa]

You haven't put a moving weight on a scale. In fact, so far, no one has suggested a way that is possible. I don't see a way that it is. Various ways to get to effective gravitational mass of a rapidly moving object (other than putting it on a scale) were suggested, because of this impossibility. Different alternatives produce different answers.My argument is simply that deflection in a rocket or 'near uniform' gravity per se, does not tell you anything about mass (because all masses will have same deflection in this setup). Maybe if you supplement this with other arguments you can get somewhere.

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.

 

[pervect]

You haven't put a moving weight on a scale. In fact, so far, no one has suggested a way that is possible. I don't see a way that it is. Various ways to get to effective gravitational mass of a rapidly moving object (other than putting it on a scale) were suggested, because of this impossibility. Different alternatives produce different answers.

I don't see the theoretical difficulty in putting a moving weight on a scale, but I do see an issue with how I described the results, which I need to revise.

My initial approach was to put an accelerometer on the moving weight, and ask what acceleration it measures. This is the proper acceleration of the weight, which,will be independent of the observer. If the acceleration measured by a non-sliding weight on Einstein's elevator is "g", the proper acceleration of the sliding weight should be $\gamma g$ by my calculation.

I don't see a problem with this calculation, but it might not be what one means by "the weight of the block". What if we wanted to put a scale on the floor and ask what it's reading was?

We do need to define what we mean by putting a scale on the floor. At speeds considerably less than light speed, this is a commercial technology, used to measure the weight of trucks without having them stop. Google for "weight in motion scale", for instance http://wimscales.com/

I'm not sure how practical it is to carry out this measurement at relativistic speeds. If we tried to use the same "load cell" technology that the truck weight-in-motion sensors use, the load cells will be slow to respond. And we have to remember that we don't have any rigid objects.

However, the end goal is just to measure the pressure on the floor. This is a well defined physical quantity, even if it might be very hard to measure in practice. The theory behind the measurement (measuring the pressure) is the same as it is for the systems that measure the weight of moving trucks. To convert the pressure to the weight, we also need to measure the footprint area of the sliding block, then multiply pressure by area.

Using this definition, we find that while the proper acceleration of the block is $\gamma m g$, the measured weight via our idealized weight-in motion pressure sensor will be $m g$.

Basically, we can compute the pressure in the block frame as (invariant) mass * acceleration / area, which is $\gamma m g / A$. The pressure transforms relativistically (as part of the stress energy tensor), the result (which I'll give without lay-level explanations) is that the pressure won't change when going from the block frame to the rocket floor frame. However, he contact area with the floor will change, the Lorentz contraction of the block will reduce the area by a factor of $\gamma$ in the rocket floor frame.

This gives the same answer as the relativistic 3-force transformation law, as it should - the force decreases by a factor of gamma.

So in conclusion, using the above discussion as a definition of how our scale works, the scale on the moving object measuring its acceleration will read $\gamma m g$, but the scale on the floor, measuring the 3-force, will read just $m g$.

My argument is simply that deflection in a rocket or 'near uniform' gravity per se, does not tell you anything about mass (because all masses will have same deflection in this setup). Maybe if you supplement this with other arguments you can get somewhere.

Using time dilation and the fact that the deflection is the same in the block frame and the elevator frame because it's perpendicular to the direction of motion, we can also conclude that the scales will read differently by the time dilation factor.

Your observation does show that the whole affair is a bit circular, but hopefullly it's useful to go around the circles to compute what the various observers actually observe. We still haven't given a really good description of things from the block frame, but I've done that before (I found some of the points interesting, but from the discussion to date I gather I may not have convinced everyone of the interesting parts). Anyway, I'm not going to get into it again unless there is some specific interest.

Hopefully this explains the block-on-the-elevator case. There is a real physical difference when we have actual gravity due to a relativistic flyby, an additional factor of two that I would ascribe to the space-time curvature that is present in the flyby case that is not present in the sliding-block-on-the-elevator-floor case.

 

[jartsa]

This even applies to a light beam; a light beam reflected between mirrors which appear to be parallel and vertical in a local frame is accelerated downwards with the same acceleration as a brick.

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.

 

[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 $\chi=0$.