Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Democritus on 02/09/2008 05:42:27
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Hi Folks
My first post to your extraordinary site...so impressed...thanks.
'Shakamaker' from 'Trainorders.com' in 2001 wrote...
"BHP Iron Ore in the north west of Australia ran the worlds heaviest ( and possibly longest) train on Thursday evening.
The train was powered by 8 AC 6000hp locos and weighed in at a staggering 82,000 wet metric tonnes + tare weight of vehicles I presume for approximately 95,000 metric tonnes.
The trains length was 7.353km (4.55 miles) long."
Well, that is where I live and I remember the train well, and saw it! And the info here is correct.
I was wondering if anyone can calculate the relativistic effects on the train given a gross mass of 100,000 metric tonnes, and a length of 7.0km (7,0000m), at a speed of 75kmh (that railroad's maximum permitted speed for empty or loaded iron ore trains.)
I am interested in the mass and dimension changes as compared to when the train is at rest. I've tried the calculation myself but get really messed up with the number of final decimal places, and I don't know if I end up talking in milligrams, micrograms etc or micrometres etc.
The effects are minuscule, which is the point of the exercise. To demonstrate the little relativistic effect upon massive (by human standards) objects at everday speeds...
Thanks if you can help,
Democritus
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Hi Folks
My first post to your extraordinary site...so impressed...thanks.
'Shakamaker' from 'Trainorders.com' in 2001 wrote...
"BHP Iron Ore in the north west of Australia ran the worlds heaviest ( and possibly longest) train on Thursday evening.
The train was powered by 8 AC 6000hp locos and weighed in at a staggering 82,000 wet metric tonnes + tare weight of vehicles I presume for approximately 95,000 metric tonnes.
The trains length was 7.353km (4.55 miles) long."
Well, that is where I live and I remember the train well, and saw it! And the info here is correct.
I was wondering if anyone can calculate the relativistic effects on the train given a gross mass of 100,000 metric tonnes, and a length of 7.0km (7,0000m), at a speed of 75kmh (that railroad's maximum permitted speed for empty or loaded iron ore trains.)
I am interested in the mass and dimension changes as compared to when the train is at rest. I've tried the calculation myself but get really messed up with the number of final decimal places, and I don't know if I end up talking in milligrams, micrograms etc or micrometres etc.
The effects are minuscule, which is the point of the exercise. To demonstrate the little relativistic effect upon massive (by human standards) objects at everday speeds...
Thanks if you can help,
Democritus
m=m*gamma , where gamma = 1/Sqrt(1-v^2/c^2)
put v in m/s, and c in m/s.
L=L/gamma (in a 'stationary' ref frame)
What answers are you getting? Also be clear about what reference frame you're referring to.
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I am interested in the mass and dimension changes as compared to when the train is at rest.
Mass doesn't change. About the lenght, d = L/γ where
γ = 1/sqrt[1 - (v/c)2]
v = train's speed = 75 km/h = 20.8 m/s
L = train's lenght in its ref. frame = 7*103 m
d = train's lenght measured in the stationary ref. frame
so d = L*sqrt[1 - (v/c)2] = 7*103*sqrt[1 - (20.8/3.00*108)2] = 7*103*sqrt[1 - 4.83*10-15]
≈ 7*103*(1 - 2.41*10-15) = 7*103 m - 1.69*10-14 m
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Hello lightarrow:
Why doesn't the mass of the moving train increase relative to the stationary train? It should, according to SR.
stevewillie
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Hello lightarrow:
Why doesn't the mass of the moving train increase relative to the stationary train? It should, according to SR.
stevewillie
Because for most physicists the concept of "relativistic mass" (the one which increases with velocity) is considered obsolete and not meaningful anylonger. What is considered as "mass" is what is also called "invariant mass" and that was also called "rest mass". The name "invariant mass" means that it doesn't change with the ref. frame.
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I don't follow that bit; if you accelerate an electron to high speeds in a Betatron, you need to take account of an effective increase in its mass. A simple Cyclotron won't work. Isn't that a relativistic increase in mass? Or, should I say, can't it be explained 'in terms of' a relativistic increase in mass?
What has changed about that since I was at Uni?
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Nothing, sophiecentaur. Lightarrow is incorrect.
(P.S. I made a typo in my initial post; it has now been corrected. m=m*gamma. gamma>=1, mass increases.)
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I don't follow that bit; if you accelerate an electron to high speeds in a Betatron, you need to take account of an effective increase in its mass.
You don't need this assumption at all. Remember that F = m*a doesn't hold in relativity (whatever meaning you give to m). A simple Cyclotron won't work. Isn't that a relativistic increase in mass? Or, should I say, can't it be explained 'in terms of' a relativistic increase in mass?
Relativistic mass is actually another name for "total energy"; since we already have a neme, why give it another one?What has changed about that since I was at Uni?
I know, I found many books talking about relativistic mass too when I was at uny, but things were changing and are still changing; this concept have been used less and less among physicists. Now just a few of them use it. Among the other things (mostly read on other physics forums) I read an article (in italian) written from an italian Prof. of Theorethical Physics which enlightened me, but it's too long to translate into english. Maybe we can talk about it a piece by time.
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Hmm... well it does take more force than expected to keep a particle in a bevatron on course as it is accelerated (beyond the force required to compensate just for it's velocity).
Of course, these masses are measured in Ev and not kg, and I wouldn't swear to how they relate to each other without checking up and reminding myself.
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Hmm... well it does take more force than expected to keep a particle in a bevatron on course as it is accelerated (beyond the force required to compensate just for it's velocity).
Yes, the force required along the trajectory is F = m*a*γ3 (the one perpendicular to the trajectory is F = m*a*γ so, if you want to keep F = M*a you have to consider two different masses along the two different directions!).
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OK, I see.
But they still 'behave' as if their mass increases. I will try to be dragged into the late 20th Century - or even the 21st..
Of course, these masses are measured in Ev and not kg, and I wouldn't swear to how they relate to each other without checking up and reminding myself.
It will be E = mcsquared, won't it?
In the betatron I think the relevant F = mA involves the centripetal force provided by the magnetic field. normal to the velocity.
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If the present fashion is to consider only rest mass as mass and the added energy due to motion as a distinct energy component, I would assume that this added energy is frame specific, ie the kinetic energy of the mass at rest plus the kinetic energy of that same mass in motion. This would seem to reduce to simple Newtonian mechanics where kinetic energy is KE=mv^2/2 and the KE of the train at rest is zero. It seems there has to be a relativistic mass component, although very small, for the moving train. Do we simply add that relativistic component as energy to the KE of the moving train or do we recalculate the whole thing in terms of potential energy where the relativistic PE of the train at rest is E=mc^2 and the train in motion is some larger value. However, it seems it there can't be a larger value if SR is correct.
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To make my question more concise, if the rest mass of the train is m and the mass of the moving train is m*, then the KE of moving train is m*v^2/2 and the PE is m*c^2? This would mean that the train gains potential energy by virtue of moving. Is this consistent with SR?
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To make my question more concise, if the rest mass of the train is m and the mass of the moving train is m*, then the KE of moving train is m*v^2/2 and the PE is m*c^2? This would mean that the train gains potential energy by virtue of moving. Is this consistent with SR?
No.
m = mass (rest mass)
v = velocity
γ = 1/sqrt[ 1 - (v/c)2]
p = momentum
E = Total energy
Ek = kinetic energy
You start from these two equations:
1. E = sqrt[(cp)2 + (mc2)2]
2. Ek = E - mc2; this is the definition of kinetic energy.
so:
Ek = sqrt[(cp)2 + (mc2)2] - mc2
If m ≠ 0, then p = mvγ so:
Ek = mc2{sqrt[1 + (p/mc)2] - 1} = mc2{sqrt[1 + (v2/c2)/( 1 - v2/c2)] - 1} = mc2(γ - 1)
and also:
E = mc2γ.
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Thanks lightarrow. Gamma is the Lorentz factor {1/sqrt(1-(v/c)^2}, is it not?. If you multiply the Lorentz factor by mc^2, it seems this is equivalent to m*c^2. It appears that this is two ways of expressing the same thing, but particles in accelerators are driven by external forces (energy is added)while the train is powered by fuel which is part of its mass. I'm not sure the two situations are really comparable.s
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Thanks lightarrow. Gamma is the Lorentz factor {1/sqrt(1-(v/c)^2}, is it not?
Of course.. If you multiply the Lorentz factor by mc^2, it seems this is equivalent to m*c^2. It appears that this is two ways of expressing the same thing
The analogy is on total energy and on momentum (in the case of m ≠ 0), but it stops here., but particles in accelerators are driven by external forces (energy is added)while the train is powered by fuel which is part of its mass. I'm not sure the two situations are really comparable.s
I sincerely don't know in Australia; here in Italy most trains are electrically driven by cables. In the case of trains fuel powered, of course, the situation is different, as you point out.