SPECIAL relativity's effect on gravity

1. Nov 26, 2013

Greetings all,

Taken in the limit of a small distance where the Earth appears flat, does a stationary object (relative to the Earth) disagree with an object moving horizontally about the force due to gravity, g? (Again, take the limit of having no altitude, so g is constant.) For example, the stationary object feels g as 9.8 m/s^2. Say a bullet is traveling at .99c. I assume it feels g as 9.8 m/s^2. However, I believe the stationary object would report there's another force acting on the bullet from it's perspective (or g is not equal to 9.8). The reason is that the two would disagree on the distance between any two points A and B that the bullet moves between. The bullet doesn't feel that it falls very far because it traveled a much shorter distance. The outside observer agrees that it doesn't fall very far and therefore says there's another force acting on the bullet. I'm aware of the effects of general relativity and time dilation due to gravitational fields, but I think we can neglect this because they are at the same height (at least for some tiny amount of time). Am I right that both the bullet and the stationary object feel g as 9.8, but the outside observer disagrees that the bullet is feeling g as 9.8? (Alternatively, the bullet could say it doesn't feel g as 9.8, but the stationary object sees g as 9.8 for the bullet, I suppose.) What is this called? Is this something used to derive general relativity? (It seems too simple to really be GR to me -- no tensor horrors :-) ).

2. Nov 26, 2013

Staff: Mentor

I am not sure what you mean by "feeling" gravity. You cannot feel gravity. If you are free-falling then you are weightless, and if you are not free falling then you feel whatever non-gravitational force is causing you to deviate from free-fall.

I think that you may be asking a question more like "what is the coordinate acceleration of a Rindler observer experiencing 1 g in the x direction in an inertial frame where he is moving at .99 c in the y direction".

Last edited: Nov 26, 2013
3. Nov 26, 2013

yuiop

Unless the bullet is travelling along the ground on wheels, it will be in free fall and will not feel any proper acceleration. However it will have a notion of how fast it is accelerating downwards. Since there is a change in height that complicates things a little bit because the gravitational gamma factor is changing, but for a small change in height in the Earth's gravity this consideration will be small relative to the velocity related time dilation. In very rough terms the clock in the rest frame of the bullet(B) will be ticking slow according to a stationary observer (E) on the Earth, so E estimates that B will record a reduced elapsed time and consequently a faster falling rate than he does. Basically the proper time recorded by the bullet will be less than the coordinate time recorded by the Earth observer, and since they both measure roughly the same vertical distance, the B observer calculates the faster falling rate.

The horizontal distance is not important as we are considering the rate of falling vertically. Technically it is related to the horizontal velocity and therefore the velocity related time dilation factor, but it probably confuses things to consider anything other than the relative velocities and the vertical distance. The vertical distance is basically the same for both observers, because it is orthogonal to the (mostly) horizontal relative velocities.

Since the bullet and the Earth based observer are at approximately the same height and so are subject to the same gravitational time dilation, the GR effects can be pretty much be ignored and the whole problem can be treated as an SR problem of transverse velocity in different reference frames. However the end result is that the observer co-moving with the bullet measures the acceleration due to gravity as being greater.

Last edited: Nov 26, 2013
4. Nov 26, 2013

Thank you, that's exactly what I wanted to know. Incidentally, is the time dilation derived from the length contraction? I'm not quite sure which to apply. If you have two points I and J on the surface of the Earth in the direction of B's travel, then E and B do disagree on how far apart they are. Why do I apply the time dilation due to velocity but not the length contraction in finding the acceleration downward for B? Why not both? If they are equivalent, as I suspect, then either would suffice -- it would just depend on how I approach solving for the acceleration due to "gravity" adjustment. Hope that makes sense; these are heady topics for me, but others do seem to approach them and communicate them effortlessly, much to my amazement.

5. Nov 26, 2013

yuiop

Once you have one the other goes hand in hand with it. Usually the consequences of relativity (time dilation, length contraction, simultaneity, etc) are derived from the two postulates.

I guess you could postulate time dilation and length contraction and the relativity of simultaneity and then derive the invariant speed of light from them is you wanted to. There are various ways to do it. Probably best to get an good introductory book on relativity for formal derivations.

Yes.
Are you referring to horizontal or vertical length contraction? There is no length contraction in the vertical direction in this case.

If you are starting out learning relativity, then introducing gravity and acceleration at an early stage probably just makes things over complicated.

Consider this analogous situation in flat space (no gravity). Consider a rod of proper length L that is at rest in frame A and is parallel to the y axis. An ant moves along the rod from one end to the other in time t and at velocity u. What would the velocity of the ant appear to be according to observer B that moving at velocity v in the x direction, with v>>u? Since the rod is orthogonal to the motion of observer B the length of the rod is the same according to both observers A and B. Once you have done that we can put the gravity back in.

Last edited: Nov 26, 2013
6. Nov 27, 2013

Yes, I was getting hung up on vertical length contraction. However, I see what you mean now; it just doesn't come into play. Thanks again.

7. Nov 27, 2013

yuiop

You're welcome :)

Now consider the velocity of the ant in the above example. In ref frame A the velocity of the ant along the rod, is measured to be greater than the velocity measured in ref frame B. If the ant was accelerating along the rod, then A would measured the acceleration of the ant along the rod to be greater than the acceleration measured by B. In the original set up that included gravity, the downward acceleration measured by an observer co-moving horizontally with the bullet would be measured to be greater than the downward acceleration measured by the observer standing on the Earth, assuming they are at roughly the same altitude. We can make this analogy, because in a local enough region of a weak gravitational field, the measurements are basically the same as in Special Relativity, to a reasonable approximation which get more accurate as the localised region gets smaller. In a very strong gravitational field this localised region might be infinitesimal.

P.S. Likewise we can consider an apple falling from a tree on Earth. To an observer moving at high speed horizontally, the apple would appear to accelerate downwards slower by a factor of (1-v^2) to a reasonable approximation, relative to the acceleration measured by an observer standing next to the tree, when the velocity of the apple is negligible relative to v.

Last edited: Nov 27, 2013
8. Nov 27, 2013

I mean HORIZONTAL (in the direction of travel). At any rate, I understand now. :-)

9. Nov 27, 2013

yuiop

Yep, the horizontal length contraction does not come into play and there is no vertical length contraction in this case, so it is just about time dilation and relative velocity. I added a bit to my last post that you might not have seen, that might be of interest to you.

10. Nov 27, 2013

Yes, we posted at the same time. Thanks for your very thorough responses!

11. Nov 27, 2013

Naty1

you do have the right answer.....SR is in flat spacetime, GR is with curved spacetime and requires the Einstein Stress Energy tensor....

here is a synopsis..
{if you want the math detail, in the discussion DrGreg shows the four acceleration ditinctions between SR and GR}

Does Acceleration in SR make it equivalent to GR?

It seems SR can handle acceleration fine, but not gravity. However, an acceleration IS gravity by the equivalence principle, so what's the difference?

Donis: This is only true locally. Beyond a small local patch of spacetime, there IS a difference between acceleration and gravity: gravity requires spacetime curvature. SR can only deal with flat spacetime; to deal with curved spacetime requires GR.

Dalespam: The difference is tidal gravity. If you have no tidal gravity then you have a flat spacetime and can use SR. If there is tidal gravity then spacetime is curved and you need GR and the EFE. For an accelerating spaceship its worldline is curved--it has nonzero proper acceleration …. but spacetime itself is still flat.

I think this may be my own interpretation:
{In flat spacetime, a body with no external forces acting on it follows a straight line geodesic and experiences no proper acceleration. In curved spacetime…. a body will follow a curved geodesic while still experiencing no proper acceleration. } Mathematically, the four acceleration in each case is zero.