# The light year long wall

1. Oct 30, 2005

Ok I have had a course is general relativity, but not on special. I was wondering something though...
If you have a brick wall a light year long (when at rest) and then have it thrusted at you at a speed such that it is only 1ft long, i.e. .99999c, when would you be able to stand in front of it without being hit by the wall?
For example does it bunch together in the middle so you can stand in front of it for 6 months or maybe it bunches towards the front so you can move into its path after just a split second? The weird thing is that if you look parallel to it you should see for the whole year right? Just something I thought was fun to think about, any thoughts?

2. Oct 30, 2005

### daniel_i_l

I'm not sure, but it dosn't seem logical that you can wait any longer because of the fact that it "bunches up". Because lets say that we are only looking at the first brick of the wall, since there is no acceleration, the rest of the wall isn't relavent (right?) cause the whole wall is moving at the same speed. The front brick only contracts a little bit so it would have almost no effect on the time. You could reason like this untill you're only looking at a tiny part of the front of the wall and then the contraction wouldn't matter. On the other hand I could be totally wrong...:uhh:

3. Oct 30, 2005

### Staff: Mentor

What do you mean by standing "in front" of it? In the path of its motion? Or watching it go by?

Realize that objects are measured to contract along their direction of motion. So I don't know why you'd think you could stand in the path of a moving wall and not be struck. (Are you thinking that it contracts perpendicular to its motion? It doesn't.)

4. Oct 30, 2005

### masudr

How can you do GR without SR?

5. Oct 30, 2005

### Jimmy Snyder

I think I understand the question. If so, then I don't think it is worded very well.
A brick wall is at rest with respect to you. The length of the wall is 1 light year, and the close end of the wall is one light year away from you. So the far end of the wall is two light years away from you.

. ................................................ .WALLWALLWALLWALLWALLWALLWALLWALL.
You .........................................Near end ............................................Far end

Forgetting dynamics and rigidity, the entire wall starts moving toward you at such speed that the length of the wall is foreshortened to 1 foot. Will the wall hit you in 1 year, 1.5 years, or 2 years?

Rather than wonder about the dynamics, let the wall remain stationary while you move toward the wall. In order to keep acceleration out of it, assume that you are already traveling at full speed at the time you reach the point labeled "You" in the diagram. If the length of the wall is measured to be 1 foot, then so is the distance from you to the near end. The near end will hit you momentarily, and the far end soon after.

Last edited: Oct 30, 2005
6. Oct 30, 2005

Oh sorry I guess I studied special.
I know you will get struck when your standing in front of the wall but when will you be struck? Because the wall is contracted to 1ft, will it be immediately, or in 6 months or 12 maybe?

Thanks jimmysnyder that is a better way to put the question.

So does this mean that you can step aside from the wall for a second then move back and the wall won’t hit you? I am aware that you can't actually move aside or back in front of the wall with out hitting it, but if you can get back in its projected path will the whole wall be behind you?

I guess the part that really boggles me is the thought that if you step aside and look perpendicular to the wall you will see it passing you for 1 year.

Last edited: Oct 30, 2005
7. Nov 1, 2005

### jackle

Because it is being accelerated towards you, I think you would find that it bunched up at your end. It sounds a bit like you are thinking of a giant holding the back end and thrusting it at you like a spear. One problem with this, is that the electromagnetic forces that push the back of the wall forward travel at light speed and take a year to get to the front of the wall! The wall is obviously going to have to compress if the back end is moving forward at nearly light speed toward the front, when the front doesn't know about it yet.

8. Nov 1, 2005

### El Hombre Invisible

The length is always contracted in the direction of motion. Whether the wall passes you or heads directly towards you it will be contracted.

The answer to how soon you can step into the wall's path is simple: 1 foot / 0.99999c seconds after the front of the wall reaches you.

I'm not sure how much sense it makes to ask whether the length is contracted to the front, middle or back of the wall in general. It depends how you set up your frame. If the front of the wall is at x = 0 at time t = 0, then the back of the wall is at x = 1 foot in your frame. If you choose a different frame, and so different values of x and/or t to measure the front of the wall, the length will be contracted differently.

It ain't called relativity for nothing! How you transform co-ordinates of one frame to another depends on what t is when x' = x.

9. Nov 1, 2005

### pervect

Staff Emeritus
I think that the question here is very similar to the question in the thread

Length contraction happens which way?

As I understand it, the problem is that you have a brick wall that's 1 light year long, and you instantaneously accelerate the entire wall up to a relativistic velocity. You then ask what the time to impact is.

The basic issue is that step #1, accelerating the enitre wall instantaneously to a relativistic velocity, is much trickier than it appears.

Infinitely rigid bodies do not exist in relativity, so if you push on one end of the wall, the other end won't start to move immediately. So you can't accelerate the wall by pushing on one end of it.

Now, you could imagine putting rocket moters evenly around the wall, and starting them all upon a signal, the signal being carefully judged so that every bit of the wall starts accelerating towards you "at the same time" and "at the same rate" where "at the same time" means "at the same time in your frame of reference". This leads to "Bell's spaceship paradox". The problem is that in the frame of reference of the wall, there will be internal stresses that break the wall if you try to do this. (It's usually thought of as a string in the Bell's spaceship paradox).

There is a way to accelerate the wall towards you without generating internal stresses in the wall. This corresponds to the notion of "Born rigidity". It's discussed in the sci.physics "Rotating disk" faq

http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

What winds up happening is that the back of the wall must accelerate harder than the front of the wall in order to satisfy the condition of "no internal stresses".