
#1
Nov710, 10:25 AM

P: 1,227

My friend is an electrical engineering major and he's probably my only closest friend whom with I can talk physics and math with for fun.
We got on the subject of that new planet that was discovered, 27 light years away, whatever. Well, light speed travel is of course impossible for any craft made of mass/humans, but his reasoning was different. "At those speeds mass approaches infinity, we'd be crushed from the gravitational forces between us and whatever craft we were in." But aside from the truth in the mass increases, it's wrong. I told him that going at a constant speed it is impossible to know if you're moving or what speed your moving at because it can be arbitrarily defined in relation to anything else; IE if there are no windows and you are moving constantly it's the same as being "motionless." If we actually felt our mass increasing through gravitational forces in the craft we would know we are moving, which relativity says is impossible to know. The best answer I could give him was that anyone in the craft does not experience the mass increase and that the mass increase is only relative to another inertial reference frame. But that leaves a lot to be desired even for me.. So, anyone got a better answer? Or is it just that simple? 



#2
Nov710, 11:22 AM

P: 474





#3
Nov710, 11:41 AM

P: 1,227

Alright, so just so I'm on the right track here, while our mass does increase relative to another reference, our gravitational constant remains the same to us because gravitational constant in our reference is based on our relative kilograms/mass?
I worded that horribly but I hope someone knows what I'm trying to say. Gravity stays the same to us because gravity in our craft is still based on our relative "increased" kilograms and not kilograms on earth or whatever other inertial frame? Is this correct? 



#4
Nov710, 11:59 AM

Mentor
P: 14,456

"No speedometer"There is an even deeper reason why your friend is wrong. We don't feel gravity. We feel every force but gravity. Astronauts float around inside the space station: it is a "zero g" environment. That does not really mean there is no gravity; the gravitational force is about 90% or so what it is on the surface of the Earth. Closer to Earth, when you go on some of the more extreme of those gutwrenching amusement park rides you are experiencing zerog conditions during parts of the ride, high g conditions during other parts. The Earth's gravitational force on you, your weight, doesn't change during those ups and downs. Your scale weight, which is what you feel, changes by a whole lot. 



#5
Nov810, 11:42 AM

P: 3,967

Inertial mass increases but gravitational mass does not increase. I know that will make the purists here flinch, so I will elaborate on that with a simple thought experiment.
Consider an 4.9m high apple tree planted at the North Pole. An apple falling from this tree takes 1 second to fall according to an observer (Frame S) standing next to this tree with acceleration (a) = 9.8 m/s^2. Now consider the same scenario from the rest frame of a space traveller (Frame S') in which the Earth and tree are moving in the East direction (x) at a velocity of 0.8c. The proper mass of the Earth is M and the proper mass of the apple is m (1kg). The apple falls in the y direction (North to South) and for this experiment ignore the acceleration of the Earth towards the apple which is negligible. To the traveller the apple takes 1*sqrt(10.8^2) = 0.6 seconds to fall. Transverse acceleration transforms as a' = a/gamma^2 = a*(1v^2). The y acceleration of the apple according to this observer is a' = 9.8/0.6^2 = 3.528 m/s^2. Transverse force transforms as F' = F/gamma = F*sqrt(1v^2). The force acting on the apple in the y direction, according to the traveller is F' = m'*a' = (gamma*m)*(a/gamma^2) = (0.6*1)*(9.8*0.6*0.6) = 2.1168 Newtons while in the Earth rest frame it is 1*9.8 = 9.8 Newtons. The force equation can be restated as F = ma = m*(GM/d^2). The transformed force is F' = m'a' = (gamma*m)*(gamma^(2)*GM'/d^2), where (gamma*m) is the relativistic inertial mass (or passive gravitational mass) and M is the relativistic active gravitational mass. When we solve the last equation for M' we obtain M' = F'*gamma/m*(d^2/G) = F*m*(d^2/G) which is the same as measured in the Earth rest frame so M' = M we can conclude that relativistic active gravitational mass is invariant under transformation. This tells us that no matter how fast the relative velocity of an object is, it will not turn into black hole. The use of the term "relativistic mass" is discouraged because it makes people think that the gravitational mass increases too, but there is no problem is it realised that inertial mass and gravitational mass are not the same thing. The concept of relativistic inertial mass has its uses and it appears in the equation for momentum P = (gamma*m)*v and is required to obtain the the correct solutions for collisions of particles to conserve total momentum of the system. It also appears in the equation for the total energy of a particle E = (gamma*m)*c^2 or alternatively E = sqrt(m*c^2 + (Pc)^2) = sqrt(m*c^2 + (gamma*m)^2*v^2*c^2). Inertial mass is simply a measure of the resistance of a object to being accelerated. In the example of the apple thought experiment, one reason that the apple falls slower according to the travelling observer is because the apple has greater inertial mass and resists being accelerated by the Earth's gravitational attraction. I think another reason the concept of inertial mass is discouraged is because as I have demonstrated above, the inertial mass and gravitational mass of an object are not the same and yet the equivalence principle is often stated in the form "inertial mass and gravitational mass are equivalent", which seams to be a contradiction to my claims. I think all that is meant by the phrase "inertial mass and gravitational mass are equivalent" is that in a small closed lab, force and acceleration due to an artificial source is indistinguishable from that due to a gravitational source. The confusion brought about by the unclear exact meaning of the phrase "inertial mass and gravitational mass are equivalent" needs to be clarified and not simply ignored and swept under the carpet by claiming that there is no such thing as relativistic inertial mass. The fact that inertial mass and gravitational mass are different things is inescapable as shown by the calculations above. The maths tells the truth. 



#6
Nov810, 12:20 PM

Mentor
P: 14,456





#7
Nov810, 02:12 PM

P: 3,967

I dispute you dismissing the thought experiment on the grounds that it is in a domain where it does not apply. The errors due to the fact that a clock 4.9m meters above another clock is running slightly faster due to gravitational time dilation are negligible. In fact we can work them out. Radius of the Earth = 6378100 m. Mass of the Earth = 5.98*10^24 kg. Gravitational constant = 6.67300*10^(11) m^3/(kg s^2) Speed of light = 299792458 m/s Gravitational time dilation factor 0.0m above the surface = sqrt(12GM/((r+0.0)c^2) = 0.9999999991995510 Gravitational time dilation factor 4.9m above the surface = sqrt(12GM/((r+4.9)c^2) = 0.9999999991995520 The error due to ignoring this difference is 0.0000000000001 % The time dilation factor due a relative velocity of 0.8c is 60% and if we factor in the gravitational time dilation factor it becomes 60.000000000001 %. Saying that error of 0.000000000001 % invalidates a calculation where the major effect if 60 percent, is nitpicking on a grand scale and amounts to a smoke screen of the real issues. So saying "inertial mass and gravitational mass are equivalent" might well be true in the context it is given in, but it is probably not relevant to the calculations I have presented and if there are there are significant errors (rather than negligible errors) that invalidate the calculations, then you should be able to point them out, because the calculations have been presented clearly for scrutiny. 



#8
Nov810, 03:00 PM

Emeritus
Sci Advisor
P: 7,437

It's definitely wrong to think that if you go too fast you'll crush yourself  or turn into a black hole. There is in fact no way to determine one's velocity relative to space, this is one of the fundamental points of both SR and GR.
Gravity does NOT reduce, even approximately, into plugging the relativistic mass into Newton's equations. So, if you are in a spaceship going at .999c, or standing still, it won't matter to what you feel. That said, violating the equivalence principle is the wrong way to explain how one gets the correct results from GR in this case. The easiest approach is to point to general covariance, and say that we know GR will do the job correctly and that there will be no effect of gravity due to velocity  because we know there is no "universal spedometer" as the thread title suggests. A slightly more sophisticated approach would be to assume you have weak fields, and then use the GEM equations. see for instance http://en.wikipedia.org/w/index.php?...ldid=393977022 for an overview [add  I'm not sure what the velocity limit of the GEM approach will be, it mentions that it's valid for slowly moving particles. It should give you at least the firstorder effects]. This requires some familiarity with Maxwell's equations to be useful. It will illustrate some of the similarities between moving masses and moving charges, by drawing a formal analogy between the equations of weakfield gravity and Maxwell's equations. (They aren't quite the same equations, even in the analogy, but they are close). The last approach is to due a full, strongfield GR analysis, though the sensible first step is to say "I can work the problem in any coordinates I want, and I'll choose to work it in coordinates where the problem is simple  i.e. where the body is at rest". Interestingly enough, most people who ask this question haven't necessarily thought about how moving charges behave, at least in my experience. The electric field of a moving charge is definitely NOT spherically symmetrical, and it is NOT given by Columb's law. Aside from this, there are magnetic effects on parallel charges (which are, after all, parallel currents) as well as electric ones. Many of the same surprises await for moving charges as for moving masses, including a gravitational analogue to the magnetic forces one sees between moving charges. 



#9
Nov810, 08:01 PM

P: 3,967

Here is an modified version of the thought experiment that might make things clearer, hopefully. Imagine we have two observers (m1 and m2) each with a proper mass of 80kg and 1,000,000m apart in otherwise flat empty space. Call the line passing through both observers the x axis. Now let us say that Newtonian physics predicts they will be 1m closer together after t seconds. 1)The gravitational mass of the observers and the velocities of the observers moving towards each other is so small in relativistic terms, that the prediction for t using the full force of SR and GR equations will differ from the the Newtonian prediction by significantly less than 1% in the rest frame of the system, so using the Newtonian equations in this case will not cause a significant error. Agree or disagree? 2)The time t for the observers to move 1m closer together, as measured by clocks held by the gravitating observers will differ from the time measured by a small clock of negligible mass at the centre of the system by significantly less than 1% and so the exact location of the measuring clock is not significant in this case. Agree or disagree? 3)The time t' for the two observers to move 1m closer together as measured by a third observer moving at right angles to the x axis with a relative velocity of 0.999c is t'=t/sqrt(10.9999^2) = t*70.712466 seconds. (The third observer can be a microprocessor system with a clock and camera weighing a few grams and sufficiently far away that the mass of this observer has negligible effect on the measured system). Agree or disagree? 4)The SR prediction that the time for the two particles to come closer together in the moving observers rest frame is over 70 times longer than in the rest frame of the gravitating system, will not differ from the prediction using the full power of GR and or Heaviside/Gravitmagnetic calculations by more than say 1%. This is essentially an SR problem. Agree or disagree? Before I go any further, I need to know if you disagree with any of the above numbered statements or if you think that GR or Gravitomagnetism calculations will produce a prediction that differs by more than 1% from the SR prediction of t' = t*70.712446 seconds. If so, can you give a rough estimate of the percentage the predictions will differ by? 



#10
Nov810, 10:26 PM

Emeritus
Sci Advisor
P: 7,437





#11
Nov910, 04:08 AM

PF Gold
P: 472

A fundamental epistemologic rule is that analogies (and all knowledge is based on analogies…) are created for a given practical purpose, to play a function. Hence they are valid in as much as such function so requires, but not beyond that boundary. The dogmatic approach forgets the function and tries to apply the letter of the analogy to the last consequences, no matter how absurd they end up being. That is the case of the Prince faced with Cinderella’s stepdaughter, who has trimmed her foot to fit it into the slipper, ie she fits into the letter but not in the spirit (the pragmatic function) of the concept. However, a dogmatic Prince would cry: I have to marry her, they are equivalent for all purposes, to all effects! And if Cinderella’s father happened to force his foot into the slipper by accident, the dogmatic scientist would still mount him on his horse, marry him and in the wedding bed, when the poor man protested “I am a man!”, our blind Prince would still insist (remember the final scene of “Some like it hot”) “Nobody is perfect…”. Coming back to the point, I fully agree that all this requires some work of conceptual clarification. I am not the most qualified for this purpose, but let me make some clumsy tries. There shoud be no problem with categorising the concept of inertial mass as a relative concept (just like time or length), as long as one understands what that means: different frames get different results about the mass of a body not because the body changes as a result of the observation (it will still have the same number of atoms) but only because the different frames, in the course of their respective measurements, interact differently with such body. (In particular, if the measurement involves the acceleration of the body, which in turn includes time and the latter is a framedependent concept…) In turn, as you point out, active gravitational mass may be an invariant concept. The two concepts are thus equivalent for many purposes, but not in this particular respect. However, I am not sure that this distinction (variant versus invariant) is so clear as it should be and I am especially not sure that the clarification derives from pure mathematics. Actually, if we say, for example, that proper time is invariant, because the proper frame measures it and the other frame “calculates” it through an equation, then we could also say the same about coordinate time: one frame measures it and the other calculates it through a formula. I think that what we try to say is rather that proper time reflects what has happened in fact: a clock has ticked a number of times (ie a number of events have happened) between two limiting events. That is reality and nobody can discuss it. It is not that the proper frame is privileged, it is just that it is better placed to obtain a direct calculation, a direct measurement in this particular case (the clock in question is at rest with it). The same applies to your example. An object cannot become a black hole due to observation. Yes, of course. But not because active gravitational mass can be mathematically calculated from any frame. It is the other way round: that cannot be because it’d be absurd, it would clash with the principle of reality and, once this is settled, we rule that it is an invariant concept (if you wish) but *in the sense that* any frame other than the proper frame, if it wants to learn what has really happened, is forced to make calculations and seek the proper concept, which is the key to the solution. Because otherwise, only on the basis of algebra and without the constraint of reality, could not I perfectly argue the following?: “No, you are mistaken, you think that you are not in a black hole, but make the appropriate calculations and learn that *you are* in such and such frame, but *you are not* in this or that frame and both things are true…” But please proceed with your calculations, because apart from this small comment, I fully agree with them and think they are very illustrative. 



#12
Nov910, 01:03 PM

P: 3,967

There are are number of errors in post #5 but it is too late to edit it, so it probably best to ignore that post and start again. (I must have been excessively overtired ). I managed to track down the old thread where we discussed the relativistic submarine http://www.physicsforums.com/showpos...85&postcount=4 and retrieved this equation for the gravitational force F acting on a particle with horizontal motion, which is useful in the context of this thread:
[tex] F = \frac{GM_0 m_0}{R^2} \frac{(1Vv/c^2)^2}{(1V^2/c^2)\sqrt{1v^2/c^2}} [/tex] where: V = horizontal velocity of the massive body (M) wrt the observer, v = horizontal velocity of the test particle (m) wrt the observer, M >> m. The equation assumes a weak field and a relativistically low vertical velocity over a short distance/duration, such that R can reasonably be approximated as constant. It can be seen from the equation that when the test particle and massive body are comoving roughly in the same direction (so that V=v) that the gravitational force acting on the test particle is: [tex] F = \frac{GM_0 m_0}{R^2} \sqrt{1v^2/c^2} [/tex] so the force acting on the test particle is actually less in this case and there is no danger of forming a black hole. In fact if the constituent particles of a body such as the Earth are in equilibrium and not not collapsing in the rest frame of the body, then the forces will still be in equilibrium when the body has relative motion, because the opposite and equal forces in equilibrium transform in exactly the same way and for such a body it does not really matter if the forces increase or decrease with relative motion. However, if we have a dust cloud that is collapsing in the rest frame of the system, then the rate at which the cloud collapses is observer dependent and is dependent on the gravitational vertical acceleration of the individual particles which is given by: [tex] a = \frac{GM_0}{R^2} \frac{(1Vv/c^2)^2}{(1V^2/c^2)} [/tex] It can be seen in the above equation that the inward acceleration of the infalling test dust particle when V=v is now given by: [tex] a = \frac{GM_0}{R^2} (1v/c^2) [/tex] which is slower than the acceleration in the rest frame of the system, so again it is demonstrated that there is no risk of a moving system becoming a black hole if not becoming a black hole in its own rest frame. The above equations were formulated to preserve the equivalence principle and it can be seen that that the rate that a falling particle falls at is independent of its horizontal velocity v when the observer is at rest with the massive body (so V=0), just as Galileo said. However it should be noted that this is only true in the weak field approximation but near the horizon of a black hole where the intrinsic curvature of spacetime is significant this relation may not hold. Looking at the equations it is hard to say anything definitive about the way gravitational mass transforms, and equation probably reflects the complex relation between how acceleration and force transform due to time dilation and simultaneity issues, more than anything about the nature of mass itself. Force is complex because it is hard to define a rest frame for force. When a particle is moving relative to the Earth for example and a force is acting between the particle and the Earth, is the rest frame of the force the Earth or the particle? I think this essentially the reason for the relative complexity of the force equation given above. P.S. I noticed in the link given by pervect http://en.wikipedia.org/w/index.php?...ldid=393977022 that Heaviside devised his equations for motion in a gravitational field including a frame dragging effect in 1893 which is not only before Einstein published general relativity but predates Special Relativity too. Does this make Heaviside the true founder of relativity? 



#13
Nov910, 04:32 PM

Emeritus
Sci Advisor
P: 7,437

I don't transform forces much, especially three forces. When I have to do a transform, I convert it to a fourforce first.
A fourforce is the rate of change of the energymomentum 4vector with respect to proper time, i.e. (dE/dtau, dPx/dtau, dPy/dtau, dPz/dtau) For comparison, three force components would be dPx/dt, dPy/dt, dPz/dt We have an extra component, dE/dtau. Let's address it's value. If we start from a frame in which the object is rest, dE/dtau is zero in that frame, as as the rate at which work is done is force*velocity and the velocity is zero. If we have some force component perpendicular to a boost, the perpendicular component of the 4vector won't be affected by the boost. Thus d (p_perp) / d tau = constant. Here p_perp is the perpendicular component of the momentum. If we imagine that the boost is in the x direction, p_perp might be Py, for example. But d(p_perp)/ dt changes when we do the boost. If we were at rest before the boost, dt/dtau was 1, and after the boost dt/dtau = gamma. Thus the perpendicular component of a 3force is reduced by a factor of gamma from what it was at rest. So far we agree, I think. However, force components that are parallel to the boost transform differently, I'm not sure whether you just didn't analyze that case or if we have a different result. If we start out in a frame where the object is at rest, dE/dtau is zero. So the only thing that happens after a boost is that our fourvector component gets multiplied by gamma. But the three force component gets divided by gamma, so there is no net change in the threeforce component parallel to a boost. While this explains how covariant forces transform, it's worth noting that if we go to the electromagnetic analogy, the electric field in isolation does NOT transform in this manner. The total force on a charge does transform covariantly, of course. But knowing that th end result transforms covariantly isn't really enough to conclude all the details of how the electromagnetic force works, and I think your approach will have similar issues with gravity. 



#14
Nov910, 06:03 PM

P: 3,967

It should also be made clear that the results you give are only applicable when both the source and the target of the force are both at rest in the same frame. If they are in different reference frames, such as a particle moving at 0.9c relative to the particle accelerator it is being accelerated by, then the parallel force is greater than the Newtonian force by a factor of gamma cubed. An equation for parallel force that tries to take this into account is: [tex]F_{} = ma\left(\frac{1V^2}{1v^2}\right)^{\frac{3}{2}} [/tex] where v is the velocity of the accelerated particle relative to the observer and V is the velocity of the accelerator relative to the same observer, although I am not sure if that works in all circumstances. This aspect does not seem to be adressed in most references, so it would be interesting if anyone else has any thoughts on this. What does not really show up in my equation is the fairly common statement that two parallel beams of light do not gravitationally interact if they are going in the same direction, but do attract if they are going in opposite directions. It would be nice to see a formula that demonstrates that, but perhaps that is digressing. You might be interested in this paper that takes an electrodynamic approach to transformation of forces. http://www.physics.princeton.edu/~mc..._64_618_96.pdf 



#15
Nov2410, 05:37 PM

P: 3,967

[tex] F = \frac{GM_0 m_0}{R^2} \sqrt{1v^2/c^2} [/tex] and when v=c this goes to zero. If the test particle is moving in the opposite direction to the attracting particle so v=V the equation reduces to: [tex] F = \frac{GM_0 m_0}{R^2} \frac{(1+V^2/c^2)^2}{(1V^2/c^2)^{3/2}} [/tex] which approaches infinite as v goes to c. The infinite transverse force is not too problematic because for a particle moving at the speed of light the rest mass has to be zero making the result effectively indeterminate (0/0) for the exact case v=c so a different equation is required for photons, but it is at least demonstrated that for massive particles the transverse force goes to zero as the velocity goes to c if they are comoving and if they are moving in parallel but opposite directions the force becomes exponentially large for an increasingly short time interval. 


Register to reply 
Related Discussions  
"Iron core energy change" and "transformers vs. ohms law"  Classical Physics  14  
"methods of theoretical physics" vs "normal math classes"  Academic Guidance  6  
Difference between "Identical", "Equal", "Equivalent"  Calculus & Beyond Homework  9 