# What is the most counter-intuitive concept in the relativity theories?

1. Mar 29, 2013

### Seminole Boy

Thanks!

Also, if someone could simplify the difference between a reference frame (used today) and a reference body (used in Einstein's book) please do so!

2. Mar 29, 2013

### phinds

It may depend as much on your own point of view as it does on the facts. personally, I find the constancy of the speed of light to be about as counter-intuitive as things could possibly get. I mean, in human terms we absolutely expect velocities to simply add and it is weird that they don't when light is involved.

3. Mar 29, 2013

### bobc2

The relativity of simultaneity and its implications. And this of course is intimately related to Phinds's comments.

4. Mar 29, 2013

### Fredrik

Staff Emeritus
One very counterintuitive result is that two observers A and B can both say that "that guy is aging more slowly than me" and both be right. It seems like an obvious contradiction at first. What you need to understand to see that it's at least not an obvious contradiction, is that statements about an observer's experience are by definition statements about coordinate assignments made by a standard coordinate system associated with his motion. So "that guy is aging more slowly than me" really means "the time coordinates assigned by my standard coordinate system to points on the curve that mathematically represents that guy's motion, are increasing faster along the curve than his age".

The SR result that I find the most counterintuitive is that if you see something like a dot from a laser pointer move faster than c, and you start running after it, its speed relative to you will be larger when you're running, not smaller.

Note that this too is a statement about the assignments made by two different standard coordinate systems.

Last edited: Mar 29, 2013
5. Mar 29, 2013

### WannabeNewton

Frame dragging always manages to blow my mind. It is easily one of the coolest things in GR.

6. Mar 29, 2013

### Staff: Mentor

You already have several good answers, but I'll add one: the fact that an object can fall into a black hole in a finite time even though the Schwarzschild time coordinate goes to infinity at the horizon. If I had a dollar for every post I've made here on PF trying to explain this to someone, I'd be retired now.

Einstein basically used the term "reference body" to mean what we would now call a coordinate system (I believe he also used the term "Gaussian coordinate system" to mean this). This is just a mapping between points in spacetime (or in some region of spacetime) and 4-tuples of real numbers, which has to satisfy certain conditions (such as continuity).

A reference frame, as the term is used today, is best modeled as a frame field:

http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

A frame field is a mapping between points in spacetime (or in some region of spacetime) and sets of vectors which can be thought of as defining the "reference frame" used by a member of some family of observers at each point.

The two concepts are closely related because in order to write down a frame field for use in calculations, you have to pick a coordinate system. But a frame field has some useful properties that a coordinate system does not, and vice versa. You pick whichever one is more useful for a given problem.

7. Mar 29, 2013

### Seminole Boy

You guys are spectacular! Let's keep it rolling.

8. Mar 30, 2013

### soothsayer

Gotta agree with Phinds and Peter Donis! Those are my favorites, in SR and GR respectively. The idea that two spaceships traveling at 0.5c in opposite directions will not see each other moving at c...and the idea that an (unfortunate) astronaut could fall into a black hole in a finite amount of time but that an outside observer would never see them cross it: just fall ever slower, and slower towards the event horizon until the end of time.

I also really like the Ladder paradox, and of course, the Twin paradox, especially the bit that Frederik described.

9. Mar 30, 2013

### zbe

Can you/someone explain this in more detail?

10. Mar 30, 2013

### HomogenousCow

In curved space time, relative velocity across a distance is ill-defined.

11. Mar 30, 2013

### Fredrik

Staff Emeritus
By far the easiest way to see this is to use a spacetime diagram. Draw the world line of a laser dot with a velocity greater than c. If you know how a Lorentz transformation tilts the axes of the diagram, you will immediately see that a boost to another coordinate system can make the speed of the dot arbitrarily large, infinite, or negative in the new coordinate system.

Alternatively, we can use the velocity addition rule. I'm going to use units such that c=1. Suppose that the dot has velocity 1.5 relative to the ground, and that you're "running" at velocity 0.6 relative to the ground. We want to find the velocity of the dot in the inertial coordinate system that's comoving with you when you're running. In this coordinate system, the ground has velocity -0.6, so we need to plug -0.6 and 1.5 into the velocity addition formula:
$$\frac{-0.6+1.5}{1+(-0.6)\cdot 1.5}=\frac{0.9}{0.1}=9.$$ So if the dot moves at 1.5c relative to the ground, it moves at 9c relative to a runner who's running at 0.6c relative to the ground.

12. Mar 30, 2013

### bobc2

Fredrik, I think yours is far and away the most interesting. Here are a couple of space-time diagrams to illustrate your example. I've just presented a succession of cases with a blue guy running faster in each diagram. The successive increase in the angles, theta1, theta2, and theta3 emphasize the successive change in slope (increase in speed) for the blue guy's X4 axis. And of course the blue guy's X1 axis (the instantaneous 3-D world that blue "experiences") successively rotates counter-clockwise to maintain the constant speed of light (green photon worldline always bisecting angle between X4 and X1 for all frames of reference, i.e., speed of light is the same for all observers regardless of their speed).

In the third frame, the red laser spot is moving at infinite speed in the frame of the blue guy.

Last edited: Mar 30, 2013
13. Mar 30, 2013

### Staff: Mentor

Last edited: Mar 30, 2013
14. Mar 30, 2013

### soothsayer

Are we sure about this moving laser dot thing? Nothing is actually moving faster than the speed of light; can we really justify plugging this apparent velocity into a S-T diagram or velocity transformation?

15. Mar 30, 2013

### bobc2

Don't worry. Nothing is really moving faster than the speed of light. That is, there is no single material particle moving. It's kind of like the old sizzors example where one has a very long pair of sizzers and closes the sizzors very rapidly. The intersection of the two blades can move faster than the speed of light. But the intersection point is not some particle or object doing the moving.

Imagine how fast you could sweep a laser spot across the surface of the moon (standing on the earth with a laser pointer). The changing position of the spot does not mean that some particle has actually moved--it's clearly not one group of photons doing the moving across the surface, and moon dust particles are certainly not doing the moving.

But, yes, you can represent that lazer spot motion in the space-time diagram--just keep track of what it represents.

This is a common example (sweeping a laser across the moon's surface) I've encountered in more than one physics class, but somehow I've never encountered the example the way Fredrik presented it--running to catch up with the spot and finding it moves away even faster. Great counter-intuitive example, Fredrik!

16. Mar 30, 2013

### Fredrik

Staff Emeritus
As bobc2 said, no particle is moving faster than light in this scenario. The motion of a dot moving faster than light obviously has to be represented by a spacelike line in a spacetime diagram, like the red line in bobc2's diagrams. Those are exactly the diagrams I had in mind when I said that the easiest way to see this is to use a spacetime diagram. The only thing that should look suspicious is that I plugged in a value greater than 1 in the velocity addition formula. If you have derived that formula by explicitly verifying that the composition of two Lorentz transformations has a velocity given by that formula, you might not expect it to be valid when one of the velocities is greater than 1. But you can also derive it in a way that makes it obvious that only one of the velocities need to be less than 1:

Let S be an arbitrary inertial coordinate system. Let x be the coordinate matrix in S of any event in 1+1-dimensional Minkowski spacetime. The map $t\mapsto t x:\mathbb R\to\mathbb R^2$ is a line through 0. It represents the motion of a point-like "object", like a particle or in this case a dot from a laser pointer, that in S has the velocity $x^1/x^0$. I number the components of the matrix from 0 to 1, and 0 is the "time" component. I will denote that velocity by v. Note that $x^1=vx^0$. What we want to do is to find the velocity of this line in another inertial coordinate system S', that has velocity u in S. I will call this velocity w. So we do a Lorentz transformation of x to the coordinate system S':
$$\begin{pmatrix}x'^0\\ x'^1\end{pmatrix}=x'=\Lambda(u)x =\gamma_u\begin{pmatrix}1 & -u\\ -u & 1\end{pmatrix} \begin{pmatrix}x^0\\ x^1\end{pmatrix} =\gamma_u\begin{pmatrix}x^0-ux^1\\ -ux^0+x^1\end{pmatrix}.$$ Since Lorentz transformations take straight lines through 0 to straight lines through 0, the velocity in S' is
$$w=\frac{\gamma_u(-ux^0+x^1)}{\gamma_u(x^0-ux^1)} =\frac{-ux^0+vx^0}{x^0-uvx^0} =\frac{v+(-u)}{1+v(-u)}.$$ If S the inertial coordinate system comoving with the ground, and S' the one comoving with the runner, then u=0.6, v=1.5 and w is the velocity of the dot in S'.
$$w=\frac{1.5+(-0.6)}{1+1.5\cdot(-0.6)} =\frac{0.9}{0.1}=9.$$

17. Mar 31, 2013

### soothsayer

Oh, I'm perfectly aware that nothing is moving faster than the speed of light (as that was my point), and I know how to do the transformations. I was just surprised that we could still plug this illusory FTL motion into the equations and get meaningful, physical results out of it.