# Weight of a Relativistic Mass

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1. Dec 4, 2014

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.

2. Dec 4, 2014

### Orodruin

Staff Emeritus
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/ [Broken].

Last edited by a moderator: May 7, 2017
3. Dec 4, 2014

### 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.

Last edited: Dec 4, 2014
4. Dec 4, 2014

### Orodruin

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

5. Dec 4, 2014

### harrylin

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!

6. Dec 4, 2014

### Orodruin

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

7. Dec 4, 2014

### Staff: Mentor

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.

8. Dec 4, 2014

### DrStupid

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.

9. Dec 4, 2014

### PAllen

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

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.

Last edited: Dec 4, 2014
10. Dec 4, 2014

### pervect

Staff Emeritus
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.

Last edited: Dec 4, 2014
11. Dec 4, 2014

### harrylin

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.

Last edited: Dec 4, 2014
12. Dec 4, 2014

### DrStupid

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.

13. Dec 4, 2014

### PAllen

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).

14. Dec 4, 2014

### jartsa

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.

15. Dec 4, 2014

### pervect

Staff Emeritus
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.

16. Dec 4, 2014

### PAllen

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.

17. Dec 5, 2014

### jartsa

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.

18. Dec 5, 2014

### PAllen

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.

19. Dec 5, 2014

### 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.

20. Dec 5, 2014

### 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.

21. Dec 5, 2014

### 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.

22. Dec 5, 2014

### PAllen

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.

Last edited: Dec 5, 2014
23. Dec 5, 2014

### PAllen

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

24. Dec 5, 2014

### Jonathan Scott

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.

25. Dec 5, 2014

### 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.