Why Does a Bullet Fall Below a Laser Beam in General Relativity?

[noobphysicist]

Well, here's the scenario I am particularly curious about:

A riffle barrel and a laser point directly towards a target some distance away. Now, General
relativity says that the bullet and the light experience the same downward acceleration
during horizontal travel, yet the bullet hits the target well below the laser beam.



Is it because light exhibits less gravitational forces than material objects such as a bullet? If not, then why does this happen?

 

[yuiop]

Broadly speaking the light and the bullet fall the same distance in the same time. In you example the light arrives at the target in less time than the bullet, so it falls a lesser distance.

 

[A.T.]

Now, General relativity says that the bullet and the light experience the same downward acceleration
during horizontal travel,

That is also what Newton says. No need to invoke GR here.

yet the bullet hits the target well below the laser beam.
... why does this happen?

Because the light reaches the target faster, so it has less time to fall. Same with bullets of different speed. No need to consider light here.

 

[Dale]

A riffle barrel and a laser point directly towards a target some distance away. Now, General
relativity says that the bullet and the light experience the same downward acceleration
during horizontal travel, yet the bullet hits the target well below the laser beam.

I think that you are thinking of the equivalence principle, which can be somewhat loosely stated as that being at rest in a uniform gravitational field is the same as accelerating in the absence of gravity.

So consider what would happen if you are accelerating in a rocket, far away from any gravitational source. Suppose your barrel and laser are aligned exactly perpendicular to the direction of acceleration. You can easily figure out where the laser and bullet will hit. If you work it out you will get that the bullet will hit well below the laser beam.

GR says that the same is true in a uniform gravitational field.

EDIT: wow! double scooped!

 

[noobphysicist]

I think that you are thinking of the equivalence principle, which can be somewhat loosely stated as that being at rest in a uniform gravitational field is the same as accelerating in the absence of gravity.

So consider what would happen if you are accelerating in a rocket, far away from any gravitational source. Suppose your barrel and laser are aligned exactly perpendicular to the direction of acceleration. You can easily figure out where the laser and bullet will hit. If you work it out you will get that the bullet will hit well below the laser beam.

GR says that the same is true in a uniform gravitational field.

EDIT: wow! double scooped!

I was wondering if maybe you can expand a little on the equivalence principle and what it has to do with my example?

 

[Dale]

There are lots of good sources about the equivalence principle. Here are a few:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/grel.html
http://csep10.phys.utk.edu/astr162/lect/cosmology/equivalence.html
http://www.einstein-online.info/spotlights/equivalence_principle

About what it has to do with your example, in the OP you claimed that "General relativity says that the bullet and the light experience the same downward acceleration" which is false in general, but correct in a uniform gravitational field. Since the usual introduction to the equivalence principle is regarding a uniform gravitational field I assumed that is what you were discussing and most likely the source of your somewhat incorrect assumption in the OP.

 

[K^2]

As somebody said, the easiest way to solve this problem is by using equivalence principle. Picture this in 2D. You have a horizontal floor running along X. The floor accelerates upwards at a rate of a in direction of Y. We can take an inertial frame from which we'll consider this problem the moment the projectile is fired. Initially the projectile is some height h above the floor and it travels in the X direction at velocity v. Some distance d from origin, projectile hits a wall. This happens after some time t = d/v after the projectile is fired. In that time, the floor has traveled distance at²/2. So the projectile strikes the wall at height h - ad²/(2v²). Note that because motion in X and Y are totally independent, we don't need to worry about relativity at all, so this works even for a beam of light. It will strike the wall at height h - ad²/(2c²). And if we simply substitute a uniform gravitational field for acceleration, we have that the light "drops" from h down to h - gd²/(2c²) if the beam is fired "horizontally". The only reason the bullet drops a lot more is because v << c, resulting in a much more significant drop.

P.S. It is very important that the projectile is traveling strictly in the X direction while acceleration is strictly in the Y direction. This is what allows us to make simple use of equivalence. If you try the same approach to a beam of light curved by a planet or a star, you'll end up making a mistake by a factor of 2.