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Does mass really increase with speed 
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#1
Dec911, 10:22 AM

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If mass really increases with speed then if 2 rockets leave earth, travelling parallel and close together, at close to the speed of light, then from an Earth point of view, as their mass increases, so should their gravity and they should be drawn together. But from the spaceship point of view their gravitational attraction should be minimal.



#2
Dec911, 11:45 AM

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Relativistic mass is an outdated concept. This is yet another example of where the concept of relativistic mass confuses rather than clarifies.



#3
Dec911, 11:52 AM

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From the point of view of Earth how is it determind the mass of the ships has increased? I think it's only kinetic energy that increases, not the actual mass. 


#4
Dec911, 02:26 PM

P: 47

Does mass really increase with speed
Relativistic mass is just the historical name for the quantity:
[tex] \frac{m}{\sqrt{1v^2/c^2}} [/tex] but it is the so called restmass [itex]m[/itex] that really has a physical significance. Rest mass is the magnitude of the fourmomentum vector: [tex]m^2 c^2=\frac{E^2}{c^2}p^2[/tex] It is an invariant quantity, meaning it is the same in every inertial frame of reference (this is a different from conservation of fourmomentum: conservation of fourmomentum refers to the fact that fourmomentum, as a vector, is conserved in any interaction, as viewed in a particular reference frame; here, we are saying that mass is an invariant quantity, i.e. it is the same when computed in all possible r.f.). I think there is even a frequentlyaskedquestion post that addresses this topic, so check it out. I am not sure it is legal to talk about gravitational forces in special relativity, so I believe it is not right to say that since the masses increase (they don't!), their gravitational interaction increases too. 


#5
Dec911, 02:36 PM

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#6
Dec911, 03:05 PM

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#7
Dec1011, 08:10 AM

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ELROCH:
Thanks for clearing that up. So its okay for electrons to orbit near the speed of light around a nucleus moving at the speed of light. Einstein would not approve. 


#8
Dec1011, 08:31 AM

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#9
Dec1011, 10:23 AM

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The main problem seems not the coordinate variance of the description, rather the question seems to assume some sort of superposition, which may not hold in GR because it is a nonlinear theory. Here is an interesting comment by Yau in this spirit (bcrowell has I think made a similar point many times, not sure if it's the same, but he's who I learnt it from, if I'm not misinterpreting him): The total energy in general relativity cannot be obtained by integrating any local density along a hypersurface. The reason is that the density would depend on the first order differentiation of the metric gij . But there is a coordinate system where such quantities are zero at that point." 


#10
Dec1011, 11:10 AM

P: 5,632

Hi lowemack..
Instead of saying 'mass' increases'' let's instead say 'energy'...that is, their kinetic energy appears to increase as observed from earth....but that is illusory since neither spaceship observes such an increase. They see earth's KE as increasing not their own. So who is right...well, they all are because they have different reference frames. So that alone won't tell us what's going on. Such comparisons can be confusing. Peter's comment is the key: Another way to remember this: Any situation where you ask about a rapidly moving massive body's gravitational effect and a 'stationary' observer can be transformed to an equivalent question about the interaction between a rapidly moving observer a 'stationary' massive body. So all observations relating to a rapidly moving massive body can be answered as if the body is stationary...as if all measures are local. Local measures trump distant measures. I think this would also be accurate: (any experts??) If the spaceships have accelerated at the same rate together for the same time as they travel, they remain stationary relative to each other, their gravitational attraction remains as if they were stationary....But if one temporarily rockets ahead faster, for example, and remains ahead, now there has been relative motion between them, kinetic energy is observed for this period, and so they would have somewhat different gravitational attractions...and their paths would now vary for future travel.....gravitational curvature IS affected. And their recorded elapsed times would also differ were they to later come together and compare elapsed times. 


#11
Dec1011, 11:24 AM

P: 253

But how does an earth based observer explain it? To him, the spaceships have increased mass, so shouldn't they have increased gravitational attraction, and if not, why not?



#12
Dec1011, 12:09 PM

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Would it help to think of this in terms of some exact solution in GR? How about the LCDM model, where we say other "galaxies (local groups)" are "moving" away from us with increasing "acceleration"? Is there any way to make sense of all those terms as well as assign them "energy" and "gravitational attraction"?



#13
Dec1011, 01:42 PM

P: 351

The 'equals' sign does not indicate a 'same thing as' relationship, but an equivalence. If the real masses did increase, it would entail an irresolvable paradox: I have installed a nanogramsensitive strainbalance between the two spaceships. I observe the readout through a powerful telescope and I see as they go faster, the readout increases. But the pilots radio back and say the readout does not change. This is not the same sort of length/time 'paradox' that can be resolved on carefully reasoned relativistic grounds, it is just an impossibility. If the pilots don't measure a mass increase, there is none. It is constant. 


#14
Dec1011, 03:21 PM

P: 262

In the past I have read arguments concluding that the force between the rockets will become smaller due to the 2 magnetic fields as observed from earth. The readout should decrease according to these arguments. I’m stuck. An irresolvable paradox? 


#15
Dec1011, 05:36 PM

P: 351

Is there greater attraction between the two? Consider two simplified spaceships: an electron and a positron traveling parallel down a linear accelerator at relativistic velocities. Do their charges change? Charge, like spin, is a fixed property. As far as I know, it doesn't change with velocity, which latter between the two is zero anyway: they have the same reference frame. 


#16
Dec1011, 06:59 PM

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However, we don't have to get into that to resolve the question of whether an object being in motion changes the gravity it produces. It doesn't. 


#17
Dec1011, 07:13 PM

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#18
Dec1111, 12:11 AM

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The observable number is invariant. How it is interpreted may vary, but the number itself does not change. Nor does it change depending on whether we do the calculation in a frame where the source of gravity is at rest, or a frame in which the source is moving. Similar remarks apply to any other observable number. 


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