uncool Posted May 13, 2008 Share Posted May 13, 2008 How does special relativity affect a gravitational field? I'd think that because the radius becomes smaller and the masses become bigger, the field is magnified by a factor of gamma^4. Is that right? =Uncool- Link to comment Share on other sites More sharing options...

swansont Posted May 13, 2008 Share Posted May 13, 2008 Gravity is part of general relativity, not special relativity. If you mean kinetic energy, AFAIK the entries into the stress-energy tensor are what dictate gravity. However, a mass change is not part of that; the mass used is the invariant mass. http://en.wikipedia.org/wiki/Stress-energy_tensor Link to comment Share on other sites More sharing options...

uncool Posted May 13, 2008 Author Share Posted May 13, 2008 Can't gravity still be used in special relativity, so long as the distortion effects are ignored? That is, you assume that gravity does not affect the speed of light/etc.? =Uncool- Link to comment Share on other sites More sharing options...

swansont Posted May 13, 2008 Share Posted May 13, 2008 Technically you have to account for effects that aren't SR if your gravitational potential changes, but there are problems you can analyze that are simple additions to SR. Link to comment Share on other sites More sharing options...

ajb Posted May 13, 2008 Share Posted May 13, 2008 You could consider linearised general relativity in the background of a flat metric. You could consider the gravitation field to be an additional field and use the Minkowski metric to raise and lower indices. I think this would be the only sensible way to do gravity in SR. Link to comment Share on other sites More sharing options...

Graviphoton Posted May 14, 2008 Share Posted May 14, 2008 Or less complex, you might want to consider just learning the basics of curved spacetime, and how curvature is totally equivalant to gravity and distortions, and even matter. Then learn about the Inverse Square Law for radiation, which must include gravitational radiation, then perhaps the gravitationally warped masses of planets. Here's something to start you off... a few words on Newtons theory... the best place to start... Newton developed his work from Kepler’s Law which describes the motions of planets. By working off Kepler’s work, he found that the planetary paths where nearly circular and they were used to describe the force causing the motion. Kepler’s three laws state: 1. All planets follow elliptical trajectories with the sun in central frame 2. A line between the sun and a planet sweep out equally 3. The ratio of the cube of the radius to the square of the period of revolution is the same for all planets. This gives us [math]k=R^{3}/T^{2}[/math] He found that the force of attraction of the sun for a planet is equal to the product of mass and centripetal acceleration. [math]F_{SP}=mPaP=mP(4π^{2}R/T^{2})[/math] Because of this, Newton started to postulate an inverse-square law of attraction. When he expressed the form of the famous inverse-law, he made it form Kepler’s third law and used [math]R[/math] and m only; [math]k=R^{3}/T^{2}[/math] and stated [math]T^{2}=R^{3}/k[/math] So the force is usually given as; [math]F_{SPOH}=mP(4π^{2}2R/R^{3}/k)=mP(4π^{2}k/R^{2})[/math] Newton introduced [math]G[/math] into his equations, and the value of the Gravitational Constant has been found to be something like [math]6.67 x 10^{-11}[/math] . Mathematically, the value of [math]G[/math] was stated as, [math]F=Gm1m2/R^{2}[/math] One interesting thing is that no one actually knows how Newton arrived at value of the Gravitational Constant. Newton also showed that force is related to mass and distance. If we consider for a moment the projectile motion of a thing, we use the equations: [math]x=v_{OH}t[/math] and [math]y=v_{OV}t-½gt^{²}[/math] They are called ‘’parametric equations,’’ and they have a parameter given as t. Solving the parameter in the first equation, we should get [math]tx/v_{OH}[/math]. Substituting t in the other equation and we get a parabolic equation; [math]y=v_{OV}(x/v_{OH})-½g(x/v_{OH})^{²}= v_{OH}/v_{OV} x-½g(x/v_{OH})^{²}=v_{OH}/v_{OV} x-g/2v_{OH}^{²} x^{²}[/math] The force of gravity is what is known as the net force on a projectile. The path of a projectile is then called the parabola. Gravity is then said to be acting on the thing along the path it takes. Even though Newton knew of the equations that described free fall, none of his equations ever required them, until Einstein’s theory of relativity came around. Suppose that M was the mass of the earth and m is the mass of a building, like the legendary Leaning Tower of Pisa, the motion given by the famous [math]F=Ma[/math] equation, must abide by the rules of gravitation, [math]F=GMm/r^{²}[/math] If m under the law of gravitation is the same value as the m under the motion of gravity, then we say, [math]ma=GMm/r^{²}[/math] Which allows for m to be cancelled out completely, leaving, [math]a=GM/r^{²}[/math] Where a is acceleration. Link to comment Share on other sites More sharing options...

Farsight Posted May 14, 2008 Share Posted May 14, 2008 How does special relativity affect a gravitational field? I'd think that because the radius becomes smaller and the masses become bigger, the field is magnified by a factor of gamma^4. Is that right?=Uncool- It doesn't sound right. I find the question a bit confusing. Are you talking about the gravitational field of a large mass, and asking if the gravitational field of that mass increases if the mass begins to move very quickly? Apologies in advance, but I'll presume yes. The "active gravitational mass" of a body is proportional to the energy of that body. A stationary object has its "invariant mass", which has an E=mc² energy equivalence. A moving object also has kinetic energy. So it has more energy, and the total is known as "relativistic mass". As a result, it has more active gravitational mass, and causes more gravity. But the object would have to be moving very fast before you noticed a difference, and it would have flashed past before you could even tell. You have to get very hypothetical and think in terms of a huge rod of superdense material. If this was shooting past you at an enormous rate, you'd fall towards it faster. Link to comment Share on other sites More sharing options...

Royston Posted May 14, 2008 Share Posted May 14, 2008 I thought the OP was asking whether SR dealt with field equations, AFAIK field equations are not applicable to SR, i.e the field equations I've dealt with are vector expressions of gravity and charge et.c, but I'm yet to study SR in depth, but AFAIK, and what I've scanned through, I can't see any mention to a field with SR, but I might be wrong. Link to comment Share on other sites More sharing options...

ajb Posted May 17, 2008 Share Posted May 17, 2008 Blagojević in his book Gravitation and Gauge Symmetries discusses attempts to formulate gravity as a field on Minkowski space-time. I think people way back in the 60's and 70's wanted to do this in order to formulate a quantum theory of gravity using what was known about quantum field theory on flat space-times. I believe this only works with linearised general relativity, i.e split the metric up into Minkowski + small fluctuations. You can then treat the small fluctuations as an independent field on your space-time. You can then couple it to other fields and particles. So in this respect, I think you can deal with gravity in SR. Link to comment Share on other sites More sharing options...

Graviphoton Posted May 17, 2008 Share Posted May 17, 2008 mmmm interesting... Link to comment Share on other sites More sharing options...

granpa Posted May 19, 2008 Share Posted May 19, 2008 electric and magnetic fields of moving objects also become length contracted. I assume that gravitational fields to the same. Link to comment Share on other sites More sharing options...

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