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.