# Fast object moving along a table

1. Jan 9, 2007

### stunner5000pt

Suppose we have a rod say 10 m long when it is viewed while both observer and rod are at rest. Now suppose this rod was moving at a speed of about 0.99c. then the rod's legnth viewd from a statioanry frame appears to be

$$L = 10m \sqrt{1 - \frac{(0.99c)^2}{c^2}} = 1.41m$$

now supose this rod encounters a hole taht is 1m long.. will the rod fall through. One of my profs told me that the rod will bend into the hole?!

Does this actually physically occur ??

Does it have something to do with someting along the lines of 'there is no perfectly solid object'?? (i just came up with taht im taking wild guesses here!)

2. Jan 9, 2007

### FunkyDwarf

It will bend through because as it slows down to go throughj the hole (ie down instead of across, thus horizontal velocity decreases) the length contraction warps, in a sense. This is due to the fact that the length that is over the hole will decelerate and move down whilst the rest still moves along behind it, causing uneven length contraction.

One might argue however that this wouldnt happen with a perfectly ridgid body as the velocity of any part is equal to any other. If you negate that and include some 'time delay' i suppose it could happen.

Without the above it would depend on the angle of entry. If its along a perfectly flat table and only horizontal velocity then id say that it wouldnt fit because it would resort to original size as soon as the horizontal velocity decreased

I think
-G

PS: Good question

3. Jan 9, 2007

### stunner5000pt

one thing though
why would hte horizontal velocity decrease?? Sinc there is gravity (i forgot to mention) why wouldnt hte horizontal velocity stay constant??

4. Jan 9, 2007

### FunkyDwarf

Ah ok, i thought it was being slowed (by someo outside force) to fall down the hole instead of skimming over it.

fair point, in that case it wont, in which case im not sure why your prof said it would bend, as there is no change in the velocity of any part of the rod.

I also sort of envisaged a tipping going on as it fell down but that shouldnt reduce Vh

I shall have to have a think :P

EDIT:

Ok i think ive got it-> using the tipping idea the rod falls in say at a 45 degree angle. This means that the direction in which contraction is occuring is no longer along the plane of the rod, and so it sort of bends, i think.

5. Jan 9, 2007

### AnssiH

He/she needs to be fired. Now.

So we have a hole in a lab-frame, and a rod that is moving at 0.99c in the lab frame?

By the "length" of the hole I assume you mean we have kind of a "pocket" that is 1 meter deep? Then the question you are asking is, does the rod fit into the pocket?

Being that according to relativity, the length of the rod in the lab frame here would be 1.4 meters, the answer is "no" (unless we start talking about foam rods, but that is not very relevant to this discussion).

If the pocket was 1.5 meters long, then relativity would say "yes", but it is not about bending or about how flexible the rod is.

If you suppose simultaneity is absolute and universal, then there is no ambiguity to the geometry of any object. But relativity works by assuming simultaneity is relative (to the inertial frame), so to define the geometry you need to state where do you find its extents at a single moment from a given inertial frame.

In the case of the rod, when viewed from its own frame, the extents of the rod are found 10 meters apart at a single moment. But when viewed from the notion of simultaneity of the lab-frame, its extents are found less than 1.5 meter apart at a single moment.

In other words, if the rod is shot into the pocket, and you measure the moment its front-end hits the bottom of the pocket and the moment its rear-end enters the pocket, you would find the rear-end entered the pocket before the front-end hit the bottom (but this ONLY when you talk about the simultaneity of the lab frame. In rod's frame the front end hits the bottom far before the rear-end enters the pocket)

Now, perhaps when your professor talks about bending, he/she is referring to what happens when the front-end of the rod actually hits the bottom. I.e. the rod needs to stop into the lab-frame, and when it does, it won't fit into the pocket anymore, so what gives?

The thing is that no matter how you'd conduct the experiment, with relevant speeds the rear-end of the rod cannot receive any information about the collision of the front-end until the back-end has already entered the hole. In other words, the rear-end cannot begin to "decelerate" to lab-frame until it is far too late. Here is where you could say that ordinary "bending" might happen, but it depends on other physical properties of the rod (perhaps it shatters).

These notions about how a rod fits into a barn and how the speed of light is the fastest you can get are basically just cute little factoids about the model we call Relativity, and they don't really help you in understanding the model. Pretty much the only thing that is relevant in understanding how relativity works is how the notion of simultaneity works in it. When you really understand that, everything else will follow effortlessly.

So your professor still needs to be fired ;)

-Anssi

Last edited: Jan 9, 2007
6. Jan 9, 2007

### pervect

Staff Emeritus
If you want a more realistic problem, imagine that you have a limp wet noodle, and that it's moving at .99c as you describe.

Why is this more realistic? On the timescales being considered, the steel bar is simply not very rigid. The speed of sound through the bar is much slower than the speed at which the bar is moving. So if you want some physical insight as to what actually happens, think of what would physically with an actual steel bar, a non-rigid object is a much better approximation than imagining a rigid object.

Another way of saying this : a spring constant has units of newtons / meter = kg / sec^2. If you take a very high speed photograph of what's going on, and play it back in slow motion, the spring constant will appear much lower.

The time it will take the front end of the wet noodle to cross the hole in the lab frame is about 3 ns. By using the principle of equivalence and imagine that your gravity field is actually generated by an accelerating rocket, you can see that the rocket will move up by (1/2) a t^2, where t=3 ns, while the head of the limp noodle will follow a straight line course and won't move up at all. Therfore the head of the noodle will drop 4.5 e-17 meters relative to the rocket in this 3ns period if your rocket is accelerating at 1 g.

This will probably be tough to detect in any actual experiment because it is such a tiny distance.

Last edited: Jan 9, 2007
7. Jan 9, 2007

### AnssiH

Oh, reading pervect's reply, now I see what the test setup actually was supposed to be in this test! (I think) So to avoid confusion I should mention that I was imagining it was about a rod entering head on a pocket 1 meter deep.

If it is moving along a table then... I'm not sure what this has to do with relativity and length contraction...

And I think your professor still needs to be fired, just in case ;)

8. Jan 9, 2007

### stunner5000pt

ok i guess this is more of a question of solid state physics but why is an steel bar not very rigid at such speeds??

looking at wikipedia's definition (not the best source...)
then te object is not rigid becuase the point masses are not continuous... they are arranged in discrete packets across the steel bar??

9. Jan 9, 2007

### pervect

Staff Emeritus
I based my remarks simply on the fact that the speed of sound in the beam is so slow. Because of this, the natural "time scale" for the beam to move will be very long, and for a short event, like passing over a hole, only the mass of the beam will be important.

I'll try to present a more formal and complete argument which will hopefully be more convincing.

Model the steel bar as a distributed spring-mass system - perhaps something as in the diagram below. We've approximated the continuous spring-mass system of the bar with a lattice network of lumped masses and springs.

Code (Text):

M---M---M---M---M
.      .     .     .      .
M---M---M---M---M
.      .     .      .     .
M---M---M---M---M

The point is that the value of the spring constant k turns out to be unimportant to the solution, and that it can be reasonably approximated as zero. Replacing the springs with a spring consant of zero gives us the "wet noodle" approximation.

To check our approximation let's try and estimate how far the system would bend if the spring force were the determining factor.

If our number is much larger than the distance that we calculated for the wet noodle case, then we know our approximation was reasonable.

This is a mechanical engiennering problem. Let's look up the answer on the internet - you can probably cross-check the answer in the mech E forum, I had to look this up, so I'm not particularly familiar with the calculations here.

http://www.engineersedge.com/beam_bending/beam_bending8.htm
http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia

Let's assume a square beam with a side s and a length L Then we can write

W = load = rho * L * s^2
E = s^4/12

and the formula from the webpage for deflection is

W L^3 / 8 E I = (rho * L * s^2) L^3 12 / (8 E s^4)

which gives 1.5 * rho * L^4 / s^2 E

with rho=7860 kg/m^3 and E=200e9 N/m^2
http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html

Given L=10m and s=.5 meters, I get a deflection of .002 meters for the steady state.

Compare this to the calculated deflection of 4.5e-17 meters due to the inertia of the beam. At .002 meters, the spring supplies enough resotring force to support the weight of the beam. At 4.5e-17 meters, the actual deflection, the restoring forces due to the "springineess" of the beam, i.e due to it's young's modulus, are totally negligible.