What happens to the car in the Relativistic Car Paradox?

In summary, two observers, A and B, have different perceptions of the same scenario involving a car traveling at 0.995c and a 2 meter ditch. Due to relativistic length contraction, observer A sees the ditch as less than 1 meter and can easily drive over it. However, observer B still observes the ditch as 2 meters and sees the car as shorter than 2 meters, leading to the belief that the car will fall into the ditch. The resolution to this paradox is that synchronicity fails and the car is able to make it across the ditch due to its momentum. Further discussion involves the idea of how a car traveling at 0.995c would roll on flat wheels and the effects of gravity on the car
  • #1
Strange Quark
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0
Here is one that's been bugging me a lot:

Observer A is zooming along in a very fast car at 0.995c. The "rest length" of the car is about 6 meters. In front of the car is a deep ditch with a "rest length" of 2 meters across. The car has big tires and could drive over the ditch under normal highway speeds.

Observer B is standing beside the ditch watching the car as it zooms by.

Due to relativistic length contraction, observer A observes the ditch in front of him to be much less than 1 meter - he should barely even feel the bump.

But to observer B, the ditch is still 2m, but he observes that the car is much shorter than 2 meters, and will inevitably fall into the ditch.

What happens to observer A? Does he make it across? Does observer B see him magically float across thin air violating Einstein's postulate: "The laws of physics shall be the same no matter what the frame of reference"?

Good luck ...
 
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  • #2
As typical with these paradoxes, the answer is that synchronicity fails so that the car can actually make it across the ditch.
 
  • #3
Not to mention/ask:
why would you think that a car traveling at 0.995c would fall into a 2 m wide crevace, regardless of how small the tires were. After 2 m, gravity would pull the car downwards by an amount (1/2)(9.8 m/s2)[(2 m)/(0.995c)]2 = 0.0002 pico-meters (which is orders of magnitude smaller than the radius of the H-atom).

This is NOT a relativistic calculation, and treats gravity as a real force. The bottom line is that the gap would fly by so fast there wouldn't be any time for the car to get pulled into it.
 
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  • #4
Originally posted by Strange Quark

What happens to observer A? Does he make it across? Does observer B see him magically float across thin air..

Apparently, Quark, you've never seen the Blues Brothers in their '56 Olds (or was it a Chevy) flying across the Hudson River at 120 mph while the draw bridge was open! Nothing magical about it; its called MOMENTUM.

And a car doing .995c means LOTS of momentum to carry it to the other side of the ditch.

Creator

Now if you want a brain bender try figuring out how a car doing .995c rolls on flat wheels.
Since the 4 foot high wheels shrink only in the horizontal direction, the stationary observer sees the car 'rolling' on vertically flat wheels!? How do you explain that?
 
  • #5


Originally posted by Creator
Now if you want a brain bender try figuring out how a car doing .995c rolls on flat wheels.
Since the 4 foot high wheels shrink only in the horizontal direction, the stationary observer sees the car 'rolling' on vertically flat wheels!?
That is an interesting point, but it isn't quite that simple. If you assume that the wheels are acting normally in the car driver's frame (that is, if you assume that the car driver observes the wheels to have a 0.995c exterior tangential velocity), then, to an observer in the rest frame of the ditch, the bottom of the tire (touching the pavement) is stationary, while the top of the tire is traveling at ~1.000c, and the sides of the tire are traveling somewhere between 0.995c and 1.000c. This causes different parts of the wheel to contract differently. It would look more squished at the top, and wider at the bottom.
 
  • #6
Ok I messed up in the details.

Forget about the wheels of the car and replace the car with a block on a frictionless table and replace the ditch with a hole in the table.
 
  • #7


Originally posted by turin
That is an interesting point, but it ... It would look more squished at the top, and wider at the bottom.

I think that's a good point...I wasn't considering the tangential wheel velocity; the wheel 'flattens' somewhat triangularly; still a paradox how a stationary frame sees it 'rolling'.

Creator
 
  • #8


Originally posted by Creator
I think that's a good point...I wasn't considering the tangential wheel velocity; the wheel 'flattens' somewhat triangularly; still a paradox how a stationary frame sees it 'rolling'.
I imagine it would look like a really fast bulldozer/tank/snowmobile tread in the shape of a teardrop. It would definitely be strange. Probably never will be observed.
 
  • #9
Originally posted by Strange Quark
Forget about the wheels of the car and replace the car with a block on a frictionless table and replace the ditch with a hole in the table.
I think you might have missed the point of mine and creator's first posts. It's the same deal.
 
  • #10
Originally posted by turin
I think you might have missed the point of mine and creator's first posts. It's the same deal.

I guess I did miss the point. You didn't convince me. Sure it may only drop a small fraction of a pm, but scale the dimensions and it eventually becomes significant. But even that doesn't matter.

It doesn't matter how little it drops (we're talking about simple ideal objects here - no tires, burms, dirt ramps or draw-bridges, if it drops at all some portion is still going to hit the inside of the ditch/hole. I don't get Creator's first post either:

Doesn't matter how much momentum you have, gravity is always acting on you in this model and without a resisting force (the table) your going down AS SOON AS the resisting force is gone. It can't fly (I forgot to say that I guess). :smile:
 
  • #11


Originally posted by Creator

And a car doing .995c means LOTS of momentum to carry it to the other side of the ditch.

Momentum has nothing to do with it. How does momentum "carry it". I hope you guys are joking.

p=mv normally

and in this case

p=ymv (where y is relativity gamma)

I gave you v. and that's all that counts as Turin showed. It's going to drop just as fast no matter what m is, assuming that the relativistic mass is still a lot less than the Earth - otherwise it may even drop faster at relativistic velocities - need to think about that one ...
 
  • #12
This discussion prompts me to ask the related questions
1) Is the concept of "center of mass" well defined for a uniformly moving object and
2) If so how would a stationary observer calculate the center of mass of such a uniformly moving object (particularly when aforesaid object is moving close to the speed of light).

If I had to guess the answer based on this thought experiment (and since I don't have the time this evening to try and formulate this in a nice mathematical way-- nor the relativistic expertise), I would assume the center of mass 'lags behind' the object in motion and can even lag far enough behind so as to be literally behind the object.

Here's my reasoning (Relativists... be gentle with me):
If the block is moving fast enough, the hole looks very small... too small in fact for the block's edge and center of mass (as calculated by the block) to both be over the hole at the same time.
Hence the block will not tip, nor fall into the hole.

To the outside observer the center of mass of the block is lagging so far behind the block that there is a Wiley Coyote moment when the contracted block is entirely over the hole... but since the center of gravity and the edge are not both over the hole at the same time-- it doesn't fall

What's more interesting is if the hole is a bit wider, or the block is a bit slower and the block DOES start to tip... but I leave that analysis to the experts.

EDIT:
That also makes me wonder... does the direction of the pull of gravity look different between the block's frame and the bystander's frame? Just to be a bit more precise... if an observer on the block (I guess we should call such an observer a blockhead) erects a pole to show the direction that they feel the pull of gravity, will a pole erected by the stationary observer point in the same direction as the blocks? Will the two poles agree when the block is brought back to the stationary observers inertial frame?
 
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  • #13
Originally posted by Strange Quark
It doesn't matter how little it drops (we're talking about simple ideal objects here - no tires, burms, dirt ramps or draw-bridges, if it drops at all some portion is still going to hit the inside of the ditch/hole. I don't get Creator's first post either:
OK, you're right; this brings up a different issue.

One issue I think we should confront at this point, though, is whether or not we really want to be talking about gravity. It would probably be better to say that the block is uniformly charged, and that there is an electric field pointing downwards. No, damit, that won't be good either. Hmm, I'll have to think about this (though I probably won't think about it).




Originally posted by curiousbystander
1) Is the concept of "center of mass" well defined for a uniformly moving object and
I hadn't even considered this issue, but I'm pretty sure it is "well defined." It may have a surprising definition, though. I'd have to think about that (and I may actually do so).




Originally posted by curiousbystander
... I would assume the center of mass 'lags behind' the object in motion and can even lag far enough behind so as to be literally behind the object.
I am leaning towards disagreement with this, but it raises an interesting issue. I'm off to think about it.




Originally posted by curiousbystander
If the block is moving fast enough, the hole looks very small... too small in fact for the block's edge and center of mass (as calculated by the block) to both be over the hole at the same time.
Hence the block will not tip, nor fall into the hole.
You would think so, wouldn't you. But, you are making an assumption that you might not be aware of, but you mention the issue at the end of your post. You are assuming here that gravity is pulling the block striaght down. I don't think that it is.




Originally posted by curiousbystander
... does the direction of the pull of gravity look different between the block's frame and the bystander's frame?
I'm pretty sure it does. But this is a GR thing, not an SR thing. There is a lot of confusion/oversimplification about the distinction between SR and GR. GR, to me, seems a gazillion times more bizarre than SR. Don't let people fool you into thinking that GR is just the inclusion of acceleration into SR, because it isn't. SR already allows for acceleration. Even in SR, though, the direction of force generally changes from frame to frame. If we're only talking SR, then I suppose we would have to treat gravitation almost exactly like electromagnetism. In that case, there would be, what some people have called, a gravetomagnetic force (as well as this so called γ factor issue). The gravitational field would be a second rank tensor (that's kind of like a 4x4 matrix). The problem is that mass is not part of a 4-vector. I suppose you could play around with the momentum 4-vector as the source of gravitation. I don't know. I don't think that it is worth playing around with it just to try to resolve this paradox.
 
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  • #14
I'm pretty sure it does. But this is a GR thing, not an SR thing. There is a lot of confusion/oversimplification about the distinction between SR and GR. GR, to me, seems a gazillion times more bizarre than SR. Don't let people fool you into thinking that GR is just the inclusion of acceleration into SR, because it isn't. SR already allows for acceleration. Even in SR, though, the direction of force generally changes from frame to frame. If we're only talking SR, then I suppose we would have to treat gravitation almost exactly like electromagnetism. In that case, there would be, what some people have called, a gravetomagnetic force (as well as this so called g factor issue). The gravitational field would be a second rank tensor (that's kind of like a 4x4 matrix). The problem is that mass is not part of a 4-vector. I suppose you could play around with the momentum 4-vector as the source of gravitation. I don't know. I don't think that it is worth playing around with it just to try to resolve this paradox.
I hope this won't be too bad... hopefully we can just describe the geodesic in one frame of reference and then pull it back along the Lorentz transformation to describe it in terms of the other (this is NOT my area of expertise so please, anyone correct my errors).

Last night before dropping off to sleep I think I realized how to calculate the center of mass. How does this sound:
We let f(x,t) represent the mass distribution of the block in space and time as seen from the observor:
We call the endpts of the block x0(t) and x1(t)
Then we use the appropriate integral to calculate the center of mass... something like integral from x0(t) to x1(t) of x*f(x,t) dx divided by integral from x0(t) to x1(t) of f(x,t) dx... or if we've defined our distribution function nicely enough... it could just be integral from -infinity to infinity of x*f(x,t) dx over the integral from -infinity to infinity of f(x) dx.

Then we use the formulas relating x, x' and t and t' , and a change of variables to recast the integral in terms of x' and t'...
What disturbs me is the possibility of a dt arising during the change of variables... but class starts in 5 minutes so I'd better stop musing and run.
 
  • #15
Originally posted by curiousbystander
What disturbs me is the possibility of a dt arising during the change of variables... but class starts in 5 minutes so I'd better stop musing and run.
It should be more disturbing if no dt arose. This is just subtle evidence that what is simultaneous in one frame is not in another. I still haven't (dove/dived/diven?) into the problem yet. I'll probably get back to it on Monday.
 
  • #16
Originally posted by curiousbystander
This discussion prompts me to ask the related questions
1) Is the concept of "center of mass" well defined for a uniformly moving object and
2) If so how would a stationary observer calculate the center of mass of such a uniformly moving object (particularly when aforesaid object is moving close to the speed of light).

Great hypothesis curiousbystander. I'm no relativity expert either but you may be on to something. If the object is relativistic I don't think you can simultaneously measure the mass of the object at all points along its length (which you need to do calculate a classical center of mass). Certainly worth some analysis - wish I had more time to try and do it. If I was only a bit smarter, then it wouldn't take so long and I might have the time!
 
  • #17
Observer A is zooming along in a very fast car at 0.995c. The "rest length" of the ca

Say the occupant of the car/block does notice a small bump. This causes him to accidentally fire his M-16 aimed toward the front of the car. It's a hot round and travels at .010C. The occupant knows he is moving at .995C. He "sees" the bullet now moving toward the front of the car/block at .010. The observer on the ground also makes this observation. Is the bullet traveling at 1.005C?
 
  • #18


Originally posted by meddyn
Say the occupant of the car/block does notice a small bump. This causes him to accidentally fire his M-16 aimed toward the front of the car. It's a hot round and travels at .010C. The occupant knows he is moving at .995C. He "sees" the bullet now moving toward the front of the car/block at .010. The observer on the ground also makes this observation. Is the bullet traveling at 1.005C?

Actally the bullet is going

w = (u + v)/(1 + uv/c2)

=0.996c relative to guy on the ground.

I hope he was wearing safety goggles and hearing protection with that hot a load!
 
  • #19
Good response. Kinda like to ask that one every now and again.

Thanks.
 
  • #20
I talked to one of my professor's today about the problem. His resolution was the opposite of mine (I think it makes more sense too-- though I didn't quite follow all his reasoning). The center of mass will lead the moving object somewhat... so from the blocks perspective gravity isn't pulling perpendicular to the block anymore... the block will tip and enter the hole at a slant...
From the bystanders point of view, the block just falls into the hole...

He also brought up the interesting point that no motion is actually rigid, and information about what's happening at the front of the block can only be transmitted to the back at the speed of light... he also suggested removing gravity and imagining a giant press pushing down so that the block is sandwhiched between the press and the surface... from the bystanders view the press pushes down on the block simultaneously, but from the block point, it starts pushing at the front and works its way back.
 
  • #21
Putting It in English and Basic Terms...

Putting It in English and basic terms, THIS is what I am hearing:

What is being said is that SIZE (I.E. MEASUREMENT) is relative to the "system" in which the measurement is being taken.

The OBSERVED size of the car at light speed, is MISLEADING.

The size of the car has never changed, the size only appears to have
changed from someone that is NOT in the same "system" as the car (i.e. the system at which observer A and the car are in, near light speed).

The reverse must also be true of the hole. It appears to be smaller to
Observer A, but the APPEARANCE is misleading Observer A, because the hole is not traveling at light speed, and the size of the hole, NOR
the car, have ACTUALLY changed, ONLY the APPEARANCE of the OBJECTS, to an Observer in a DIFFERENT system, makes it SEEM AS IF the size of either object has changed.

Now, I want to mention one other thing that nobody even mentioned.
As speed increases, so should mass. As mass increases, so should the effect of gravity, upon that object.

So, if an object is traveling near light speed, shouldn't it's mass
be NEARLY infinite? (Or at least kinda heavy).

It would lead me to ask the people that were calculating the gravity thing, that the gravity should be acting upon an extremely heavy object that is just itching to fall right through the highway, and would take the soonest chance it could, to fall into any hole.
Am I misunderstanding something, anything?
 
  • #22
You can do the thought experiment at finite speeds. Here's a handy thing to remember, at 87% of c the shrinkage factor for length and time is 1/2 and the magnification factor for mass is 2. So the car will be half as long wrt the ground and the ditch will be half as long wrt the car. And yes, the car will be twice as massive wrt the ground, although that does not affect the problem.


Work on the simultenaity angle. The ends of the car are spacelike related, so simultaneous events at one end aren't simultaneous at the other end. Ditto for the dirch.
 
  • #23
selfAdjoint:

I'm intrigued by this one too -- although I think the mass, momentum & gravity aspects are beyond my reach right now.

But -- simultaneity of what? What is being assumed to be simultaneous, aside from the car being above the ditch simultaneous with the ditch being under the car (if, in the frame of the ditch, the car is shorter than the ditch)?
 
  • #24
The bystander will observe the front and back of the contracted block falling at almost the same time (that's very close to her idea of simultaneous), but an observer attached to the block will see the front end tilt and fall into the hole at an angle.. in other words, the back of the block will start falling later then the front of the block by a very perceptible amount (that's his idea of simultaneous).

That's also a response to Treat2's objection... the contraction is real. And the mass of the object isn't going to affect how fast it falls.
 
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  • #25
What would cause it to tilt? If the hole is shortened to a small fraction of the length of the block, and the block is constructed so that whenever its center of mass is over the shortened hole, both the front and back ends are supported by solid ground, why wouldn't it just slide right over?
 
  • #26
If my professor is right (and he has an annoying habit of being exactly that), and I'm interpreting what he said correctly, then from the point of view of the block, gravity is not pulling at an angle perpendicular to the block, but rather from further forward and down. The faster the block is moving, the more pronounced the effect-- so the front edge of the block AND the center of gravity will both be 'over' the hole at the same time (realizing that the wall of the hole are no longer parallel to the pull of gravity, and the angle of the 'floor' is no longer perpendicular to that pull either)
 
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  • #27
So ... even though the block is bigger in its frame than the size of the (contracted) hole, it "squirts" in like a cartoon character squeezing into a bottle?

Bizarre!
 
  • #28
Originally posted by curiousbystander
If my professor is right (and he has an annoying habit of being exactly that), and I'm interpreting what he said correctly, then from the point of view of the block, gravity is not pulling at an angle perpendicular to the block, but rather from further forward and down. The faster the block is moving, the more pronounced the effect-- so the front edge of the block AND the center of gravity will both be 'over' the hole at the same time (realizing that the wall of the hole are no longer parallel to the pull of gravity, and the angle of the 'floor' is no longer perpendicular to that pull either)

There was an article published in the American Journal of Physics by Wolfgang Rindler a long time ago which discusses this. I think this problem was called the manhole paradox or something like that.
 
  • #29
Originally posted by pmb_phy
There was an article published in the American Journal of Physics by Wolfgang Rindler a long time ago which discusses this. I think this problem was called the manhole paradox or something like that.
One paradox published by Rindler around 1961 was called the "Paradox of the Fast Walker" I think. That might be the one you're thinking of: it is very similar to the one discussed in this thread.
 
  • #30

1. What is the Relativistic Car Paradox?

The Relativistic Car Paradox is a thought experiment that explores the effects of time dilation and length contraction as predicted by Einstein's theory of special relativity. It involves a car traveling at near-light speed and a stationary observer, and raises questions about the perception of time and space in different frames of reference.

2. How does the car in the paradox appear to move?

From the perspective of the stationary observer, the car appears to move at a speed close to the speed of light. However, from the perspective of the car's driver, the car appears to move at its normal speed. This is due to the effects of time dilation, which causes time to appear slower for objects moving at high speeds.

3. What happens to the length of the car in the paradox?

According to the theory of special relativity, objects in motion will appear to be shorter in the direction of motion when observed from a stationary frame of reference. This is known as length contraction. Therefore, the car in the paradox will appear shorter to the stationary observer than it does to the driver.

4. Does the car in the paradox experience any changes in time?

Yes, due to the effects of time dilation, time will appear to pass slower for the car's driver compared to the stationary observer. This means that the driver will experience less time passing during their journey than the observer. This phenomenon has been confirmed through experiments with high-speed particles.

5. Is the Relativistic Car Paradox possible in real life?

While the thought experiment is a useful tool for understanding the principles of special relativity, it is not possible for a car to reach speeds close to the speed of light in reality. The energy required to accelerate an object to such speeds is currently beyond our technological capabilities.

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