# Do objects appear heavier if they run over a scale very fast

• I
Hi.

I've just re-read a high school introduction about SR. It introduces the relativistic mass. I know that this concept isn't used anymore in modern formulations of SR, but observations should be the same in all formulations. They make the following thought experiment: An electric tram and a battery powered motorbike both drive over a scale at speeds close to ##c##. They then argue that only the tram will appear heavier on the scale since it receives its kinetic energy through electricity from the power lines from above, it's an open system. The motorbike though is a closed system, it transforms the energy saved in its battery to kinetic energy, so its total energy and hence mass remain constant.

Is this correct? I remember that the relativistic mass is a subtle issue and one needs to distinguish between longitudinal and transverse mass (which they don't in this text). Wouldn't we need transverse mass here (since gravity works perpendicular to the direction of motion)?

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.Scott
Homework Helper
That is basically correct. Of course, if the bike has reached relativistic speeds solely on battery power, that is one special bike.

• pervect and Sorcerer
That is basically correct. Of course, if the bike has reached relativistic speeds solely on battery power, that is one special bike.
Nah, don’t be fooled by the charlatan Sorcerer’s trick. See, the battery powered bike reached a relativistic speed by hitch hiking on a spaceship, then was released right before riding over the scale. Of course the rider and bike were obliterated shortly after due to friction.

pervect
Staff Emeritus
Hi.

I've just re-read a high school introduction about SR. It introduces the relativistic mass. I know that this concept isn't used anymore in modern formulations of SR, but observations should be the same in all formulations.

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They make the following thought experiment: An electric tram and a battery powered motorbike both drive over a scale at speeds close to ##c##. They then argue that only the tram will appear heavier on the scale since it receives its kinetic energy through electricity from the power lines from above, it's an open system. The motorbike though is a closed system, it transforms the energy saved in its battery to kinetic energy, so its total energy and hence mass remain constant.

Is this correct? I remember that the relativistic mass is a subtle issue and one needs to distinguish between longitudinal and transverse mass (which they don't in this text). Wouldn't we need transverse mass here (since gravity works perpendicular to the direction of motion)?

Let's overspecify the problem to hopefully make the concepts a bit clearer. We have a bike that is 50kg empty unfuelled, and has 50kg of fuel. Let's assume the fuel is matter and antimatter (25 kg of each). We use total conversion technology to convert the fuel with 100% efficiency into electrical power to power the 100% efficient electric motor. And we assume there is no friction or air resistance. IT's a good idea to do the experiment in a vacuum anyway, meteorites an spacecraft moving at velocities that are much much slower burn up in the atmosphere all the time.

If the bike were only fusion powered, one would have to deal with what happened to the fusion byproducts. Most likely one wouldn't be able to reach such a high velocity. Wiki estimates maybe 10 percent of the speed of light for a fusion powered spaceship, just for a ballpark figure - I haven't done the math. With total conversion, and 100% efficiency for everything, we don't have to worry about this issue. Any sort of chemically powered motor or battery would not be able to reach a significant fraction of the speed of light.

The same sort of effects should happen in chemical reactions that do in nuclear and matter-antimatter reactions, by the way. The point is that they're so small that we don't (AFAIK) have direct experimental confirmation of what one might call "the change in mass". We do have some direct experimental confirmations on stronger, nuclear reactions. You can look up the rest masses for such nuclear reactions in wiki, and confirm that the rest masses of the reaction products are lower than the rest mass of the reactants - though not by such a large 2:1 factor as in our thought experiment with the matter-antimatter powered bike.

We calculate the resulting velocity of the bike: it's gamma factor is ##\gamma = 1/\sqrt{1-v^2/c^2} = 2##, by the conservation of energy. To confirm the conservation of energy, we note that the rest energy of the bike was 100kg * c^2 initially, and the total energy of the moving bike is 50 kg * gamma, with gamma=2. A gamma factor of 2 works out to velocity v = ##\sqrt{3}##/2 = .866c.

So, we have a fueled bike that weighed 100kg to start. It has a rest mass of 100kg and the scale reading of the bike at rest was 100kg. Assuming no energy losses due to inefficiency or friction, we wind up with a bike with a rest mass of 50kg. But the scale reading on the truck scale for this now-unfuelled bike is 100kg.

I haven't belabored the semantic issues, but some clarification is needed. Rest mass is a property of a body that's independent of it's state of motion. It's the sort of mass most physicists use. Physicists typically just call it "mass".

The weight on the scale is not the rest mass, the weight on the scale depends on the energy of the object that's moving over the scale. The energy of the object is not just a property of the object itself, because it depends on the speed of the object, and the speed of the object depends on the frame of reference.

Ibix
It occurs to me that anything doing a significant fraction of lightspeed is well above orbital velocity under most circumstances. Will it weigh anything at all?

• Sorcerer
PeterDonis
Mentor
The motorbike though is a closed system

No, it's not. If it's on the Earth, it is exchanging energy and momentum with the Earth by pushing on the Earth with its wheels. If it's a spaceship out in space, it has to have a rocket exhaust to accelerate, which means energy and momentum is being transferred to the exhaust. Either way it's not a closed system.

PeterDonis
Mentor
We calculate the resulting velocity of the bike: it's gamma factor is ##\gamma = 1/\sqrt{1-v^2/c^2} = 2##, by the conservation of energy.

What about the energy (and momentum) that had to be transferred to the Earth for the bike to accelerate? (If we were considering a spaceship, it would be energy and momentum transferred to the rocket exhaust.)

Baez's article on the relativistic rocket equation is a good resource for the spaceship version of the problem:

http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

jbriggs444
Homework Helper
What about the energy (and momentum) that had to be transferred to the Earth for the bike to accelerate?
As for a classical car on traditional highway, the energy transferred to the Earth [as judged in the center-of-momentum frame] is negligible as long as the resulting motion of the highway surface is negligible. Work = force times distance. The Earth will not move significantly in response to a 50kg motorbike being given 50kg of mass-equivalent energy. One hopes that the highway surface is rigidly attached.

They make the following thought experiment: An electric tram and a battery powered motorbike both drive over a scale at speeds close to ##c##. They then argue that only the tram will appear heavier on the scale since it receives its kinetic energy through electricity from the power lines from above, it's an open system. The motorbike though is a closed system, it transforms the energy saved in its battery to kinetic energy, so its total energy and hence mass remain constant.

The result is basically correct for mass, but that's not what the scale would measure under the assumed conditions.

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PeterDonis
Mentor
As for a classical car on traditional highway, the energy transferred to the Earth [as judged in the center-of-momentum frame] is negligible

Negligible compared to the total energy of the Earth, yes. Negligible compared to the total energy of the car, in a case where we are allowing the car to achieve relativistic speeds, no.

jbriggs444
Homework Helper
Negligible compared to the total energy of the Earth, yes. Negligible compared to the total energy of the car, in a case where we are allowing the car to achieve relativistic speeds, no.
Sorry, I am not following. If you are merely allowing a 50kg motorbike to accelerate using 50kg of energy, the energy of the recoiling earth would be negligible compared to the kinetic energy of the motorbike.

PeterDonis
Mentor
If you are merely allowing a 50kg motorbike to accelerate using 50kg of energy, the energy of the recoiling earth would be negligible compared to the kinetic energy of the motorbike

Ah, I see--the word "recoil" rebooted my brain... • phinds
pervect
Staff Emeritus
What about the energy (and momentum) that had to be transferred to the Earth for the bike to accelerate? (If we were considering a spaceship, it would be energy and momentum transferred to the rocket exhaust.)

Baez's article on the relativistic rocket equation is a good resource for the spaceship version of the problem:

http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

As far as the momentum and energy transfered to the Earth, the total momentum needed to be transported to the bike will be ##\gamma m v## = 2 * 50kg * .87 c = ##2.6*10^{10} ## kg m / sec. The Earth has a mass of about ##6*10^{24}## kg, so the recoil velocity of the earth will be ## 2.6 * 10^{10} / 6 *10^{24}## = ##4*10^{-15}## m/s, where we can use the non-relativistic formulae because the recoil velocity of the Earth is so low. The energy lost in this process will be ##.5*m_e*v_e^2##, which is about 47 microjoules by my calculation. So it's not a big factor, considering that the bike has a kinetic energy of 50kg * c^2, about ##5*10^{18}## joules, a little over the energy released in a 1000 megaton nuclear explosion according to the www.

Additionally, the bike will have to accelerate at a very high proper acceleration in order to reach the specified velocity before it travels a distance greater than the diameter of the Earth. I think it'd be easier to drill a hole through the Earrth so the bike could travel in a straight line than make the bike follow a curved path on the surface, but I could be wrong about that. One could use the formula that you cite from Baez cites to figure out the required proper acceleration, I'll simply note that it's "unreasonably high". For a thought experiment, it might be better to imagine a 1 lightyear racetrack in empty space, though I don't know what to do about the gravity needed to hold the bike on the road in that case.

Don't get me started on the requirements of the tires :).

I think the poster who said "one very special bike" had the right idea.

• jbriggs444
Mister T
Gold Member
It occurs to me that anything doing a significant fraction of lightspeed is well above orbital velocity under most circumstances. Will it weigh anything at all?

You are using the word weight to refer to a force.

However, in the message you responded to that same word seems to be used to refer to the mass.

So, we have a fueled bike that weighed 100kg to start.

It really is best to use the word force when referring to a force and mass when referring to mass. :)

The word weight has different meanings in different contexts. Merchants use it to refer to what physicists call mass, and that is consistent with the way the laws governing commerce are written. Physicists usually insist that the same word refers always to a force, but don't seem to be able to agree on just how that force should be defined. Some insist it's ##mg## where ##g## is the free fall acceleration and some insist it's ##mg## where ##g## is the gravitational field strength, the difference being due to Earth's spin.

• SiennaTheGr8
The result is basically correct for mass, but that's not what the scale would measure under the assumed conditions.

When I read it citate egain I realized it's the other way around in regard to the mass:
Neglecting the increase of the internal energy of the tram (e.g. by heating, rotating parts, etc.) it's mass will remain constant but it's energy will increase.
Neglecting loss of energy (by friction, EM radiation, etc.) and the ransfer of the energy to the ground (e.g. by assuming the mass of the ground to be infinite) the energy of the bike would remain constant but it's mass would decrease.

The problem with the measurement remains. How about assuming the ground to be an infinite plane? It would have an infinite mass, the gravitational field is homogeneous and the trajectories of the vehicles are streight lines. Is it possible to calculate the readings of the scale in this special case?

PAllen
There are results like the following, which suggest a factor approaching 2 might be needed, if speed is close to c.

https://arxiv.org/abs/gr-qc/9811052

Note, this paper first analyzes beams of material dust.

PAllen
A different way to look at the question, which avoids all discussion of fuel, is to ask, given a frictionless object, will identically constructed scales moving under it at different speeds read different weights, and by how much? If the scales are also frictionless in their motion over the earth, and their read speed is arbitrarily fast, and all is done in a large vacuum chamber, many complications are removed. My guess is that the scales will read approaching 2 gamma times rest weight when moving very near light speed.

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pervect
Staff Emeritus
If the moving object generates gravity / curves spacetime, there is justification for a factor of 2:1 in the amount of curvature/amount of gravity generated by an object approaching the speed of light. But that's not the problem being asked - we can do the problem in flat space-time on Einstein's elevator with special relativity, for instance, and see there is no 2:1 factor.

PAllen
If the moving object generates gravity / curves spacetime, there is justification for a factor of 2:1 in the amount of curvature/amount of gravity generated by an object approaching the speed of light. But that's not the problem being asked - we can do the problem in flat space-time on Einstein's elevator with special relativity, for instance, and see there is no 2:1 factor.
But is that valid? Since energy gravitates, any object sufficiently fast enters the GR realm. If you assume massless with no KE, then there is no weight in the first place. I guess, though, you could ask the question under the assumption that GR doesn’t exist, only SR exists.

But then this raises interesting questions about the POE, since this experiment can be made very local, and there are no charges involved!

PeterDonis
Mentor
If the moving object generates gravity / curves spacetime, there is justification for a factor of 2:1 in the amount of curvature/amount of gravity generated by an object approaching the speed of light.

No, it isn't, because speed is frame-dependent. The source of gravity is the stress-energy tensor, and you have to include all of the stress-energy in the problem, including the stress-energy that accelerated the object. There are exact solutions in GR that model idealized versions of this kind of case: for example, the Kinnersley photon rocket and the Axelburg-Seixl ultraboost (I think I've spelled the latter correctly). Unfortunately, I can't find good online references for either of these right now (there used to be some and I'm pretty sure there are past PF threads linking to them). Neither of these solutions looks anything like "gamma times the amount of gravity the object would have at rest ".

A different way to look at the question, which avoids all discussion of fuel, is to ask, given a frictionless object, will identically constructed scales moving under it at different speeds read different weights, and by how much? If the scales are also frictionless in their motion over the earth, and their read speed is arbitrarily fast, and all is done in a large vacuum chamber, many complications are removed. My guess is that the scales will read approaching 2 gamma times rest weight when moving very near light speed.

What we have now is a special relativity 'paradox'.

The scale says that the attractive force between the tram and the earth is reduced by gamma compared to the rest force, because of the motion of the tram-earth system.

The people standing on the earth say the same thing about the scale.

(Let's say the scale is an inflated air mattress. The people say that the pressure inside the mattress has not changed compared to the rest pressure, but as the mattress has contracted, the force on the contracted surfaces is reduced by gamma compared to the rest force)

(Or maybe the scale consists of two positive point charges, in this case the people might say that the magnetic attraction reduces the net force between the charges. (By gamma))

(Yes, I applied special relativity's force rule on a gravitational force. If two masses exchange amount x of momentum gravitationally, and it takes y seconds of time in the rest frame of the masses, it takes two times more time according to a guy traveling past the masses at speed 0.86 c. Right?)

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Ibix
You are using the word weight to refer to a force.

However, in the message you responded to that same word seems to be used to refer to the mass.
True. But the diagram in the OP is apparently using a spring balance, so is measuring weight. And if we're talking mass then the answer is trivial: invariant mass does not change, relativistic mass (standard caveats about the term apply) does.

@pervect's idea of doing this in an elevator in flat spacetime seems sensible. In that case, the force applied to a mass with velocity ##v## such that it undergoes proper acceleration ##g## perpendicular to the velocity is ##\gamma mg##. So a mass ##m## is "heavier" by a factor of ##\gamma## than it is at rest, at least in the sense of the reading on a naive spring balance, I think. Edit: that answer's PAllen's no fuel, no friction variant, anyway. You need to work out the rest mass of the accelerated motorbike/tram to do the original problem. That suggests to me that the moving motorbike will be lighter than the tram (assuming their initial mass was equal), the moving tram will be heavier than the stationary one, and the motorbike weight will depend on its fuel efficiency.

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Is this correct? I remember that the relativistic mass is a subtle issue and one needs to distinguish between longitudinal and transverse mass (which they don't in this text). Wouldn't we need transverse mass here (since gravity works perpendicular to the direction of motion)?

They are talking about the transverse relativistic mass, because relativistic mass is the transverse one, unless explicitly said to be the longitudinal one.

The longitudinal relativistic mass of an object is not related to the total energy of said object, the transverse relativistic mass is.

I mean, if burning 1 kg of fuel can increase the longitudinal relativistic mass of the motorcycle by 100 kg, then the longitudinal relativistic mass can't be the gravitating mass.

And if we're talking mass then the answer is trivial: invariant mass does not change, relativistic mass (standard caveats about the term apply) does.

In case of the bike the exact opposite is true.

PAllen
So, to answer the question I posed re-the POE (principle of equivalence) versus the factor of 2 that appears in some GR scenarios:

All of the scenarios in GR where the extra factor of 2 appears (typically, 1+β2 is the actual factor, being exactly 2 for light) are inherently non-local. One is measuring total deflection of a test body by a passing mass, with the whole trajectory contributing to the deflection, and comparing to Newtonian expectation. In the case of beams, one is measuring deflection over a finite distance and comparing this to a Newtonian model where one assumes just the total energy density of the beams are the source of gravity. Again, you find an extra factor of 2. One can say these are both coupled to the impossibility of an exact SR inertial frame over a finite region in GR (in the case of beams, you can talk about gravitomagnetism augmenting versus cancelling attraction, for parallel versus anti-parallel; again, this effect is not a local effect visible in a neighborhood of a single event). Note, in particular, the analysis of beams uses very long beams.

Thus, for the case of frictionless ball moving relativistically over a small, instant reading, scale, the POE should apply, and the weight should be γm, both for a rocket or sitting a planet. By construction, the acceleration is orthogonal to velocity of the ball, so we have the pure transverse relativistic force formula.

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But the diagram in the OP is apparently using a spring balance, so is measuring weight.

It actually measures the force than prevents the vehicles from following their geodesics. I don't know if weight is a useful term in this context but in order to calculate it you need to know the trajectory of the vehicles. I don't see how it is defined in the description of the thought experiment.

Edit: quote corrected

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PAllen
It actually measures the force than prevents the vehicles from following their geodesics. I don't know if weight is a useful term in this context but in order to calculate it you need to know the trajectory of the vehicles. I don't see how it is defined in the description of the thought experiment.
You screwed up your quoting. What you quote is by Ibix, not by me. I've discussed only a much simpler, alternative scenario to the one in the OP.

pervect
Staff Emeritus
But is that valid? Since energy gravitates, any object sufficiently fast enters the GR realm. If you assume massless with no KE, then there is no weight in the first place. I guess, though, you could ask the question under the assumption that GR doesn’t exist, only SR exists.

But then this raises interesting questions about the POE, since this experiment can be made very local, and there are no charges involved!

For a formal justification of what I'm trying to say, look at https://www.physicsforums.com/threa...elativistic-speeds.948690/page-3#post-6008592 and compare it to the papers you cite. This is just one of a number of posts/threads on the topic.

The short summary is that we analyze superman flying (running, or riding a motorcycle , or skating on roller skates, or just sliding across the floor would also all work) on an accelerating elevator in flat spacetime. We don't see any factor of 2 in the proper acceleration. "Superman" can be replaced with the motorbike - or a sliding block, as has been done in other threads. Since the latest incarnation is a motorcycle, let's use "motorcycle" for consistency, though past posts have used other analogies.

In this analysis, we see a factor of ##\gamma^2## for the proper acceleration of the motorcycle. There's no factor of 2.

This post doesn't analyze the weight of the cycle in the lab frame, it analyzes something else. So we need to convert the results to get the weight in the lab frame.

This may be slightly tricky, but basically what supports the motorcycle is some pressure (force/unit area) P with some contact area A. The prouct of this pressure times the area is what we actually measure when we weigh the moving bike with a truck scale.

What happens is that the pressure P scales as ##\gamma^2##. It's perpendicular to the direction of motion, so the pressure (a component of the stress-energy tensor) is the same in the lab frame as it is in the motorcycle frame. The contact area of the tires, however, is not the same in both frames. The contact area in the lab frame is smaller by a factor of ##\gamma## due to length contraction, thus the weight of the bicycle, the product of the pressure * contact area, only scales as a factor of ##\gamma##. There is no factor of 2.

The weight of the bike is frame dependent, because it's the amount of momentum transfered per unit time. Due to time dilation, a unit time on the bike is not the same as a unit time in the lab. The amount of momentum transferred in a unit of proper time is a frame-independent quantity, the amount of momentum transfered in a unit of coordinate time is not.

The paper you (Pallen) cite are not wrong, but they're analyzing a different problem. Analyzing the problem in the flat space-time of special relativity is the difference, and it's why we see a factor of 2 in one case and not the other.

It's worth checking to make sure that I'm analyzing the same thing the OP is asking. I believe I am.

Hi.

They make the following thought experiment: An electric tram and a battery powered motorbike both drive over a scale at speeds close to ##c##. They then argue that only the tram will appear heavier on the scale since it receives its kinetic energy through electricity from the power lines from above...

It might be interesting to re-work the problem in the curved space-time of GR, the analysis I did above showed no factor of 2 in the flat space-time of special relativity. I think the key feature in the GR version of the problem there will be how we determine the shape of the floor. In the presence of acceleration and/or space-time curvature, "flatness" is frame dependent, so saying that the floor is "flat" needs a bit of clarification to be unambiguous.

Hopefully we haven't confused the OP too much, but I wouldn't bet on it :(. Apologies for bringing A level material into an I/B level thread.

T

PAllen