# Relativity of length contraction

1. Jun 28, 2012

### coktail

If two clocks with observers attach to them move away from each with symmetrical acceleration, they will experience symmetrical time dilation, meaning they both see the rate of each other's clocks slow down relative to their own. However, if the acceleration is asymmetrical, the time dilation will be as well (e.g. Clock A will see clock B slow down, and clock B will see clock A speed up).

Stop me if I'm wrong here, but I think that's correct.

Does this same principal apply to length contraction, or is it always perfectly symmetrical?

As always, thank you.

2. Jun 28, 2012

### GAsahi

It is not correct.

3. Jun 28, 2012

### harrylin

That's rather incorrect:
You cannot determine the rate of a "moving" clock with only a single "stationary" clock. For that you need a reference system with either multiple clocks or assumptions or both, with assumptions often different from the way you put it. However, the comparison of the total recorded time lapse when such clocks are brought together is just as you say.
The clocks register accumulated "time". That's different from length which is more like clock frequency: for an object in equilibrium those are independent of what happened in the past.

4. Jun 28, 2012

### GAsahi

He's not talking about "bringing the clocks together".

5. Jun 28, 2012

### harrylin

Exactly, that 's why I corrected his/her post.

6. Jun 28, 2012

### coktail

Clearly I have gotten in the way of my own question. Sorry about that. Let me try again.

Two astronauts float in space, astronaut A and B. They accelerate away from each other with symmetrical acceleration at .9c. Both A and B report that the other astronaut appears contracted and "skinny.: Correct?

Is there an scenario in which one reports the other as contracted/skinny, but the other does not? Or, a case where one reports the other as appearing skinny, but the other as expanded? I'm pretty sure there's no such thing as length expansion, but I thought I'd throw that in there for good measure.

7. Jun 28, 2012

### GAsahi

Length contraction is a function of relative speed. Both astronauts measure each other as being contracted by the same amount, $\sqrt{1-(v/c)^2}$.

8. Jun 28, 2012

### Staff: Mentor

There is not. Both see the other one as contracted, as both see themselves as at rest and the other one moving relative to them.

If they stop moving relative to each other (say they decelerate, turn around, return to some common location and stop relative to each other so that they can compare notes) they will both see the normal rest lengths and they will both agree that their clocks are now ticking at the same rate.

If all their movements while they were apart were symmetrical, they will also agree about how much time they spent and how much distance they traveled on their journeys. If the movements were not symmetrical, they may not agree about this, but once back at rest with each other they will still see both clocks ticking at the same rate and both lengths the same. This is roughly analogous to the way that the speedometer will read the same (zero) for two cars parked side by side, but depending on how they got to the parking spot the odometers may (and generally will) read differently.

9. Jun 28, 2012

### Darwin123

One can include acceleration in a "special relativity" type theory merely by differentiating the equations of the Lorentz transformation with time. This mathematical approach could also be considered "general relativity" rather than "special relativity". However, this would be semantics.
To some scientists, general relativity is special relativity after the the equivalence principle is hypothesized. To other scientists, general relativity is special relativity with any term for acceleration. Regardless, acceleration can be incorporated into the Lorentz transformation merely by differentiating the equations.
One physical interpretation of this mathematical approach is to hypothesize that a physical reset of the clocks in a frame of reference can't occur while an external force is being applied to the clocks. This physical assumption leads to the asymmetry of observers in accelerating (i.e., noninertial) frames.
Let me apply this idea to length contraction/ length dilation. The Lorentz transformation of the x coordinate between two inertial frames (S and S') is"
x'=(x-vt)/√(1-v^2/c^2)
t'=(t+vx/c^2)/√(1-v^2/c^2)
where v is the S' frame relative to the S frame, c is the speed of light in a vacuum, etc. Differentiating of these two equation will explain most of the apparent "asymmetries" in special relativity. One can use this approach to explain how a time dilation with acceleration becomes a time contraction. However, I will try to explain the question being asked by the OP>
I am doing this off the cuff, without any proofing or editing. Feel free to find my mistakes, if any. I once did this before, but I may have forgotten some of it. The important thing is that in the "twin" conundrum, if a<0 and v> 0, then dx'/dx>0. A large acceleration toward earth makes the effect into a "length dilation" rather than a "length contraction".
Now S' will be considered the accelerated frame and S will remain an inertial frame. Physically, this corresponds to applying an external force on the observer in the S' frame. Therefore, both sides of the equation will be differentiated by x. The relativistic length correction, with both general and special relativity taken care of, is dx'/dx.
1) dx'/dx=(1-tdv/dx-vdt/dx)/√(1-v^2/c^2)-(dv/dx)(v/c^2)(x-vt)/(√(1-v^2/c^2))^3
Chain rule (Calculus)
2) dv/dx=(dv/dt)(dt/dx)=a/v
where a is the acceleration. Also, by definition of velocity.
3) dt/dx=1/v.
Substitute 2 and 3 into 1:
4) dx'/dx=(-ta/v){1/√(1-v^2/c^2)}{1-(2v/c^2)(x-vt)/(1-v^2/c^2)}
where a is the dynamic acceleration. In other words, set
5)a=F/m
where F is the external force on the observer in the S' frame, and m is the mass of that observer.
Let us consider the case of the where the "space ship" is turning around, so that x=vt and a<0. If x=vt is substituted into equation 4:
6) dx'/dx=-(ta/v)/√(1-v^2/c^2)
Since a<0 and v>0.
7) dx'/dx>0.
When the twin starts his turn around, with a large acceleration, he sees his earth twin stretched out rather than contracted.
Not that the asymmetry comes about not due to the kinematics, but due to the dynamics. The asymmetry in the problem comes in equation 5. For the accelerating twin, in the rocket, the external force is nonzero (i.e., F>0). However, the "earth" twin who isn't in the rocket experiences a zero external force (i.e., F=0).
The external force prevents the clocks from resetting. With that extra hypothesis, special relativity becomes self consistent. Note: there is no way to resolve the asymmetries in special relativity without some extra hypothesis. There is no way to make special relativity "self consistent" with acceleration unless one in some manner takes into account the forces that move the observer.
I like the force hypothesis (equation 5) because it is easy to analyze. What I do when I start a special relativity problem is draw the force diagram on the hypothetical observer for each part of the trip. I let F be the total external force on whatever observer I am analyzing. I differentiate the Lorentz transformations, apply simple calculus, substitute equation 5 into the differentiated equations and then analyze what happens.
Try it with dt'/dt. You will see how the "twin paradox" is not a paradox.

10. Jun 28, 2012

### coktail

I'm sorry to be such a layman, but what I take from this is that you are saying that length dilation IS possible? This seems contrary to everything else I've read, including this thread. Now I am confused.

Thanks for taking the time to respond.

11. Jun 28, 2012

### GAsahi

Unfortunately, you cannot apply the above transformations in the case of accelerated frames. You could use Rindler coordinates or the equations of hyperbolic motion but not the above transforms as a valid starting point. While your approach is interesting, it isn't valid. There are a few more problems with the derivation, see below.

dx'/dx>0 doesn't mean length dilation, it simply means that dx' and dx have the same sign.

Nothing validates x=vt at the turnaround.

dx'/dx>0 doesn't mean "stretching".

12. Jun 29, 2012

### Darwin123

Then you haven't read very much about general relativity. What you read was restricted to special relativity. For instance, there is a problem in general relativity regarding the ratio between the circumference of an circular orbit and the diameter of the circular orbit. The ratio is less than π. In other words, the circumference has undergone a "length dilation".
General relativity deals with the issue of "asymmetry" more directly than special relativity. Two observers in different gravitational potentials determine distances and time intervals differently. The asymmetry in gravitational potential leads to an asymmetry in measurement. In unmodified special relativity, the source of the asymmetry is less obvious than the source of asymmetry in general relativity.
To discuss the asymmetries of special relativity clearly, we have to both know the Lorentz transform equations. The Lorentz factors for length and time are only valid under a very specific type of measurement. Although the Lorentz factors are taught first, one has to be aware of the complete Lorentz transform.
You have to understand the difference between an inertial frame and an accelerating frame. In the strictest sense, special relativity only with measurements made in inertial frames. Observers in two inertial frames are not accelerating relative to each other. However, neither observer in an inertial frame is accelerating in the dynamic sense. Therefore, each will observe "the length" of the a ruler in the other frame as being shorter than the corresponding ruler in its own frame.
My point is that the "correction factor" for length is dx'/dx and the "correction factor" for distance is dt'dt. If both frames are stationary relative to each other, so that v is constant, then:
dx'/dx=1/√(1-v^2/c^2)
dt'/dt=√(1-v^2/c^2)
However, the expressions for dx'/dx and dt'/dt become more complicated if v is not constant. If v is not constant, then there is a force applied to the clocks and rulers of one of the frames. In the frame where force is applied, the acceleration (a) is not zero.
If you have taken introductory calculus, then I suggest the following. Determine the expressions for dt'/dt and dx'/dx using the rules of differentiation. First do it with the assumption that v is constant in time. Then do it with the assumption that v varies with time. Look at the resulting expressions for dt'/dt and dx'/dx. Then, make your own physical interpretations based on the expressions for dt'/dt and dx'/dx.
Here are two equations for the Lorentz transformation. If S is traveling at a velocity v in the x direction with respect to frame S', then,
x'=(x-vt)/√(1-v^2/c^2)
t'=(t-vx/c^2)/(1-v^2/c^2)
Using these Lorentz transform equations and the rules of differentiation, one can determine the expressions for:
dx'/dx = ???????????
dt'/dt = ???????????
Discussions about special relativity would be so much easier with these expressions posted.

13. Jun 29, 2012

### Darwin123

You said,
"dx'/dx>0 doesn't mean length dilation, it simply means that dx' and dx have the same sign."
You are correct. Sorry. I meant that "stretching" (length dilation) is
dx'/dx<1.
The dx' is an infinitesimal length as measured in the S' frame. The dx is an infinitesimal length as measured in the S frame. If S' is the rest frame, then dx'/dx is the ratio of length measured in the rest frame to the ratio of length measured in the
Therefore, dx'/dx is the ratio of the length measured in the rest frame to length measured in the moving frame. If this ratio is less than one, then the
You said that,
"Nothing validates x=vt at the turnaround."
I was referring to the twin conundrum. The word "twins" implies that the two observers were at the same place at the same time moving at the same velocity at least once in their lives. Therefore, their clocks were "reset" at the time of takeoff, t=0. If I define t is the time in the S frame where the S' twin turns on his rockets to go home.
I may be getting my primed and unprimed coordinates mixed up. However, I once did it more carefully.
How would you physically interpret dx'/dx and dt'/dt?

14. Jun 29, 2012

### GAsahi

You are still faced with the fact that your starting point is incompatible with what you want to achieve. Your starting point should be the equations of hyperbolic motion (look up how Bell's Paradox is solved on wikipedia). You will find out that there is no such thing as length "dilation", even in the case of accelerated motion.