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If an object was sliding on an infinitely long friction-less floor on Earth with relativistic speeds (ignoring air resistance), would it exert more vertical weight force on the floor than when it's at rest?

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    $\begingroup$ The question has to make explicit how the answer is supposed to treat the effect of the curvature of earth's surface. $\endgroup$ – Johannes May 9 '13 at 15:34
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    $\begingroup$ @Johannes Or we could just assume a hypothetical universe in order to answer the spirit of the question $\endgroup$ – Jim May 9 '13 at 15:35
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    $\begingroup$ I am not a relativist and I hesitate to write an "answer", but my understanding is something like The "relativistic mass" is energy and all energy couples to gravity, so the short answer is "yes", but there are some caveats because it also shows up in other elements of the stress-energy tensor generating contributions with the opposite sign so that you can not, say, accelerate a body until it forms a black hole. Not sure what the final effect is. $\endgroup$ – dmckee May 9 '13 at 15:41
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    $\begingroup$ Interesting question $\endgroup$ – Andrew Palfreyman May 14 '13 at 19:28
  • $\begingroup$ The answer is just yes. $\endgroup$ – Ron Maimon Aug 22 '13 at 22:52
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First off, your question is phrased in terms of relativistic mass, which is an obsolete concept. But anyway, that's a side issue.

The question can be posed in terms of either the earth's force on the puck or the puck's force on the earth. We expect these to be equal because of conservation of momentum.

In general relativity, the source of gravitational fields is not the mass or the mass-energy but the stress-energy tensor, which includes pieces representing pressure, for example. The puck has some stress-energy tensor, and this stress-energy tensor is changed a lot by the puck's highly relativistic motion. Therefore the puck's own gravitational field is definitely changed by the fact of its motion. However, the change is not simply a scaling up of its normal gravitational field. The field will also be distorted rather than spherically symmetric. Yes, the effect is probably to increase its force on the earth. The earth therefore makes an increased force on the puck.

Here is a similar example that shows that you can't just naively use $E=mc^2$ to calculate gravitational forces. Two beams of light moving parallel to each other experience no gravitational interaction, while antiparallel beams do. See Tolman, R.C., Ehrenfest, P., and Podolsky, B. Phys. Rev. (1931) 37, 602, http://authors.library.caltech.edu/1544/

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    $\begingroup$ "Two beams of light moving parallel to each other experience no gravitational interaction, while antiparallel beams do." Ahhhh... This general relativity thing is still too difficult for me... $\endgroup$ – Calmarius May 9 '13 at 18:38
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    $\begingroup$ I wonder how close the anti-parallel beams have to be in order to orbit each other. $\endgroup$ – Thomas May 9 '13 at 20:58
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"Does gravity depend on relativistic mass or rest mass?" is a rather interesting question -- Einstein's initial approach was to say "the relativistic mass", and this was the pre-general relativistic answer, but this is not satisfactory, since the relativistic mass is only one component of the energy-momentum vector (and it would actually sound more reasonable to say it depends on the rest mass).

This is the theoretical motivation for general relativity, in which gravity depends on the stress-energy-momentum tensor, and only the time-time component of gravity, $\Gamma^i{}_{00}$, depends on the relativistic mass, whose density is also the time-time component of the stress-energy tensor, and the other components depend on on the other components of the stress-energy tensor by the Einstein field equations.

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    $\begingroup$ Minor quibble: Any reason to avoid refer to GR as level 3 approximation? If level infinity is taken, where does that leave a future QG theory? $\endgroup$ – Johannes Jul 20 '13 at 7:50
  • $\begingroup$ The actual level 2 approximation does not change r in a straightforward way, it has an attraction given by the product of the energies (relativistic masses) but with a peculiar radial dependence given by the gravitomagnetic contribution, which is best to work out by boosting to the rest frame of one of the objects, where the gravitational field is known. $\endgroup$ – Ron Maimon Aug 22 '13 at 22:53
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My previous answer proved to be wrong. Energy density ("relativistic mass") does contribute to gravity - and the fact that the object is moving at relativistic speeds does affect the space-time around it.
There is an interesting document that explains the problem in further context.

Besides, when we think about it, if energy density didn't contribute to the force of gravity, photons, mass of which consists of their energy density, would not be affected by gravity. But we have provided many times, that the lightwaves - thus photons - are affected by gravity.

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    $\begingroup$ It's okay to say that the object's rest mass remains the same and classically gravity depends on rest mass. But, don't we consider mass-energy? (which increases exponentially with velocity). I think it still affects the stress-energy tensor. $\endgroup$ – Waffle's Crazy Peanut May 9 '13 at 15:33
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    $\begingroup$ I'm just reading some works that actually prove that my answer is wrong. I'll edit it. $\endgroup$ – Tomáš Zato May 9 '13 at 15:35
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    $\begingroup$ Gravity does not couple to the rest mass. $\endgroup$ – Johannes May 9 '13 at 15:36
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    $\begingroup$ Gravity does not couple to the rest mass. - That would mean that still objects are not affected by the gravity - wouldn't it? After reading a few documents, I'm quite concerned that gravity is affected by the total $m$ which equals to: $\frac{m_0}{\sqrt{1-v^2/c^2}}$ so the rest (or invariant) mass affects gravity too. I might be confused by the terminology though. $\endgroup$ – Tomáš Zato May 9 '13 at 15:46
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    $\begingroup$ What I mean is that gravity couples to energy and momentum (and their fluxes) and not to rest mass (which your earlier answer stated). $\endgroup$ – Johannes May 9 '13 at 18:49
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Assuming the frictionless floor is locally horizontal, orthogonal to the radial direction and ignoring the effects of the spinning earth, the force on the floor would actually be zero when you reach the velocity of $v=\sqrt{GM/r}$­ where r is the radial distance to the center of the earth. This is because the sliding object would now be in orbit.

I suspect the force on the sliding floor generally could be written as:

$$F=-\frac{mGM}{r^2}\hat{r}+m\frac{v^2}{r}\hat{r}$$

If you have pure non-radial motion. This would become zero long before relativistic effects become noticable. I guess a totally correct answer depends a bit on if the floor "bends with the Earth" so it is always orthogonal to the radial direction. The answer here is correct if the floor "bends with the Earth".

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protected by Qmechanic Aug 9 '13 at 4:45

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