# Is "length contraction" an illusion?

1. Jun 5, 2014

### BiGyElLoWhAt

I've read a lot of places about length and time dialation or length contraction of an object relative to an observer, but what I haven't found is whether the space that the object occupies is actually being altered. It makes more sense to me that it would only appear as though the length changed.

I suppose my explaination for this phenomena would be as follows:

Imagine an object of 'rest length' L_r and an observer a distance R_b from the base of our object, distance R_c from the center of our object, and distance R_t from the top of our object at time t_observer = t_object = 0. (I'm assuming for simplicity that the times are sychronized in the rest frame).

If we accelerate both our observer and our object at a rate of a=a_0, then (since we base size off of the recieved light), we can say that our inertial length (sorry if I'm botching these terms, I'm pretty new to these concepts, hence this thread) is going to be the same as it was in the rest frame, assuming that L_r is negligibly small compared to R_c, as our object and observer are moving at the same velocity and can thus be considered at rest with each other. This length should be

...

I have to go now, I'll finish my post when i get home from work. #No save as draft function?

 Part 2:

Ok, firstly, I want to amend my situation slightly. I want to put my observer on top of our light source, giving him the ability to switch it on and off, and also a way to measure time immensely accurately.

Call $P(t)_{top}$ the position of the top of our mobile object and let $P(t)_{top} = P(0)_{top} + \dot(P)t$ (I'm assuming that the whole object moves as one, with the same velocity, and it is thus unnecessary to label the velocities Ptop and Pbottom)

Call $P(t)_{bot}$ the position of the bottom of our mobile object and let $P(t)_{bot} = P(0)_{bot} + \dot{P}t$

Call $X(t)$ the position of our mobile observer and let $X(t) = X_{0} + \dot{X}t$

looking at the case of a stationary object and observer, we can see that the time it takes for a "ray of light" to leave our observer and hit the top of the object is
$\Delta t_{top} = \frac{P(0)_{top} - X(0)}{c}$
and bottom is
$\Delta t_{bot} = \frac{P(0)_{bot} - X(0)}{c}$
(remember our observer is starting and stopping the clocks)

Thus the "rest length" of our object would be $c \Delta t_{top} -c \Delta t_{bot}$

Now let us give our objects velocities.

If we focus in on the top of our object, we can see that the time it takes light to leave our observer, hit the top of our object, and once again arrive at our observer will be equal to

$\Delta t_{top} =$ ...
OK this ended up being a lot more tedious than I thought.

I have solved for the distance 1 way (light leaves the observer and reaches the object)

$\Delta t_{top_{on\ the\ way\ there}} = \sqrt{\frac{P(0)_{top} - X(0)}{c^2 - \dot{P}^2}}$

It might end up taking me a while to solve for the time it takes to get back and hit the observer, but once I get that solving for the bottom will merely be a matter of substitution.

taking $P(t)_{top} = \dot{P}(\Delta t_{top_{on\ the\ way\ there}} + \Delta t_{top_{on\ the\ way\ back}})$ and subtracting $P(t)_{bot}$ which will be a very similar equation, we will get the illusion of length dilation, (I presume)

I solved this using pythagorean's theorem with the base being equal to $P_{top_{0}} - X_{0}$, the height being equal to $\dot{P}\Delta t$ and $\Delta t$ being equal to the hypotenuse (distance traveled by the light to intersect with an object moving at $\dot{P}$ ) divided by c. Solve for $\Delta t$

The way back won't be as simple because I have to account for the velocity of my observer, which adds another unknown.

I think it is apparent what I am going for, and through this method I will arrive at (at least conceptually, not necessarily mathematically) the same conclusion of length dilation as relativity (I'm not sure if it's GR or SR).

Can I get some input?

Please note that I'm not trying to discredit the relativities, merely understand what they are saying by providing how I'm interpreting the phenomena, and compare and contrast the two.

Last edited: Jun 5, 2014
2. Jun 5, 2014

### phinds

Time dilation and length contraction are an effect of remote observation.

You, as you are reading this, are traveling at .999999c from the frame of reference of an accelerated particle at CERN and from that frame of reference you are massively length contracted. Knowing that, do you feel any shorter?

3. Jun 5, 2014

### WannabeNewton

It actually is altered. In relativity, the concept that space is altered based on the choice of reference frame is almost a tautology. A lot of the conceptual issues people seem to have with length contraction can be easily ameliorated if one learns relativity at the outset using Minkowski's geometric language. J.L. Synge appealed to a statement made by Minkowski in his (Synge's) wonderful GR book, and I paraphrase, that "relativity" was the wrong qualification for this and the general theory; the theories are in the end about absolute Lorentzian space-time. That space and time separately are entirely relative notions is an immediate consequence of this (I am brushing aside a mass of details of course). If space itself is a relative notion, relativized to different reference frames, then of course space and the relative measurements of lengths will vary from frame to frame. Calling this an illusion would be a gross misrepresentation of what the beautiful, geometrical, space-time formulation of SR has to say about how different observers foliate space-time into different families of spatial slices at given instants of their clock time.

4. Jun 5, 2014

### BiGyElLoWhAt

@WannabeNewton

Then is there no cap to the amount of dilation that occurs (assuming c is not necessarily the speed limit of the universe (I know that it's postulated in one of the relativities, and I see the logic, I'm just trying to see if I can reach a similar conclusion from scratch)?

I'm at a very elementary understanding of GR and SR, but to me it seems as though you could achieve the illusion of length contraction but looking at the velocity, not the speed, of light relative to your observer, and the difference in time it takes light from one end of an object to reach the observer compared to the amount of time it takes light from the other end of the object to reach the observer; i.e. the distances are either a) different, or b) changing at different rates relative to our observer.

I'm going to try to finish my post, I haven't formalized this concept mathematically before, but I should be able to do so using relatively simple geometry along with some kinematics.

5. Jun 5, 2014

### D H

Staff Emeritus
If it's just an illusion, how do you explain how muons with their extremely short 1.56 microsecond half life can go from the upper atmosphere to the surface of the Earth? From the perspective of an Earth-fixed observer, the explanation lies in time dilation. From the perspective of a muon-fixed observer, the explanation lies in length contraction. Time dilation and length contraction are flip sides of the relativistic coin.

6. Jun 5, 2014

### pervect

Staff Emeritus
My \$.02. Most people who ask "Is length contraction an illusion" are presuming that length is real. I am assuming the Origial Poster falls in this class. But what is "real"? It's an ambiguous term, much debated by philosophers, and in general it's hard to tell what is meant by "reality, making it hard to answer questions about.

Let me say what I think the issue is more precisely. It is the question "Is length independent of the observer". Observer-independence is one of many qualities that "real" objects and quantites are supposed to have. I think it is this particular aspect of "reality" - observer independence - that is the underlying question we need to ask about length here.

The resolution is simple, though difficult for many to accept. Length is NOT independent of the observer. So length contraction becomes a non-mystery, because we don't expect two observers to necessarily measure the same length. It does turn out that two observers with the same velocity measure the same length in flat space-time, but observers in relative motion are a counter-example to the idea that any two observers will measure an object to have the same length.

7. Jun 5, 2014

### D H

Staff Emeritus
I agree. I have a similar problem with questions that ask if the curvature of space-time is real. Curved with respect to what? Apparently it's with respect to the One True Physics™, that of Newton.

8. Jun 5, 2014

### BiGyElLoWhAt

Firstly, post amended (still not finished)

Secondly, @DH and Pervect,

I'm not attempting to say that length is independent of the observer,

Maybe my real problem is that I don't understand what 'Length Dilation' is meant in the context of relativity. Will you please see my amended post and respond accordingly?
(I really don't feel like working through the math to finish solving this problem, as it has proven more tedious than expected)

9. Jun 5, 2014

### pervect

Staff Emeritus
If you really want to work through a particular example, a diagram would help a lot.

Quoting what you wrote so far:

I don't have a clear picture of where the light source is in relation to the object, which direction the object is moving - i.e. if I try to draw a diagram from your verbal description, I'm not sure how to start drawing it.

However, I would like to put a plug in for NOT drawing your own diagram or writing your own description, but stealing one from a textbook example of the length contraction problem. I'd suggest Ben Crowell's SR textbook, except I *still* can't get it to download :(.

Amongst other things, the idea of having things accelerate is going to complicate the problem. Standardized examples are good because you can check to see if your answer is right, and because they've been designed to be as simple as possible without extraneous complications.

I'm not clear if you are familiar with the Lorentz transform. If not you are not likely to stumble on a correct derivation of length contraction, though it is possible (for instance using the Bondi approach using doppler shifts.)

Speaking of which - Bondi's book (Relativity and Common Sense) seems to be online in an archive at https://archive.org/details/RelativityCommonSense, you can find the derivation of the Lorentz transform and of the Lorentz contraction given the Lorentz transform on pg 117-121.

10. Jun 5, 2014

### BiGyElLoWhAt

I thought I designed this to be fairly simple. It's 1 dimensional motion that I had in mind, although I don't feel that I designed my equation to be limited to 1 dimension. I will draw a picture, though, more or less because I think it would be REALLY cool to try to work through this without any formal training.

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11. Jun 5, 2014

### BiGyElLoWhAt

I assumed no rotation, no acceleration, i.e. I looked at the situation where they were stationary and I looked at the situation where the object moved at $\dot{P}$ and our observer moved at $\dot{X}$

I don't see how it can get much simpler than that.

12. Jun 5, 2014

### BiGyElLoWhAt

I took a ray of light leaving our observer (and the light source) at t=0 (he starts his stopwatch) and they began moving at the velocities given. I then calculated the time it takes the ray of light to leave the observer at t=0 and reach our object on it's path when it's moving at Pdot (I actually calculated how long it takes for the light to hit the TOP of the object, as we will eventually need the time it takes for a ray to travel from our source to the top and our source to the bottom)

Does this help at all? I know it's rather minimalistic, but that's what we do, right?

13. Jun 6, 2014

### pervect

Staff Emeritus
The textbook case is considerably simpler. Everything is in line, so it's a one dimensonal problem in space, you only need an X coordinate to specify the position of an object. (Plus you need a time coordinate).

Here is a diagram of the setup (first figure) and of a radar measurement of the distance to the end of the rod (second figure). Note that in the second figure, the times T1' and T2' are measured on the moving observer's clock (hence the prime symbol), while T3 and T4 are measured on the stationary obsever's clock.

We make use of the fact that if a radar signal is emitted at time Ta and received at time Tb, the position of an object at time (Ta+Tb)/2 is equal to (Tb-Ta)/2c.

We also make use of the fact that at T=0, T'=0 and X=0 for the moving observer and one end of the rod whose length we are measuring. Thus we only have to measure the position of the other end of the rod (the one not at the origin of our diagram) according to the moving observer.

If the length of the rod in the stationary frame is L, and the length of the rod in the moving frame is L', given the constancy of the speed of light we can then write

T2' -T1'= L'/2c
T4 - T3 = 2L/c

Note that order to find the position of the end or the rod at T'=0, we additionally need that T1' = -T2', because of the radar equation mentioned previously = the time of measurement is Ta+Tb/2 thus we must have Ta = -Tb.

This gives us that

T2' = -T1' = L'/c

Using the fact that that the doppler shift factor, which will will call K, will be constant and equal between the two observers, as discussed in Bondi's book, we can write:

T1' = K*T3
T4 = K*T2'

The doppler factor is the ratio of the recieved frequency to the transmitted frequency or the transmitted period to the recieved period. If we moudlate every n'th transmitted pulse to make it stand out be being more intense (amplutide modulation), because of the 1:1 correspondence between received and transmitted pulses, we can say that any signal sent at time T by the transmitter clock is receive at a time KT by the recieved clock, when the clocks are moving away, and by a time (1/K) T when the clocks are moving towards each other, hence the above equations.

Given L', we know T1' and T2', and we can solve for T3 and T4

T3 = (1/K) (L'/c)
T4 = K (L'/c)

and find that L = (c/2)*(T4-T3) = L' * (K+1/K)/ 2

We thus see that L is always greater than or equal to L', being equal only when the doppler shift factor K is equal to 1 which implies no relative motion.

Or to put it another way, L' is Lorentz contracted.

Using the results from Bondi's book, we can look up K as a function of velocity, which is derived in Bondi's book by finding the inverse function , V as a function of K, using a similar radar setup to the one we used here.

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14. Jun 6, 2014

### Staff: Mentor

Its purely an effect of geometry like rotating a rod to fit through a door - in fact the analogy is formally the same because length contraction can be described by hyperbolic rotation.

Thanks
Bill

15. Jun 6, 2014

### PAllen

Fine, as long as you apply it equally to time dilation. Time dilation is different (coordinate) 'cross section' for time like interval under hyperbolic rotations. Length contraction is different cross sections of a spacelike interval hyperbolic rotations. For the physics (as real as it gets) of muons reaching the ground from the upper atmosphere, the reason the muons reach the ground in the earth frame is hyberbolic rotation of the timelike interval of the muon's life (cross section increases); in the muon frame, it is due to hyperbolic rotation of a spacelike interval through the atmosphere (cross section decreases).

Further, you can get the analog of differential aging (two clocks differ when separated and brought together through different world lines) for distance using an idealized odometer. The observers meeting again will disagree on 'amount of travel' accumulated on their respective odometers. A muon in a ring will measure the amount of lab 'road' gone by as small; the lab will measure the length of ring comoving with the muons (muon road) as much larger (over one circuit). They compare their odometers after one circuit and they differ, inversely to the difference in their clocks.

16. Jun 6, 2014

### BiGyElLoWhAt

I guess I'm not seeing how you can Length dilation if your object is separated by 1d space in 1d time.

At any given time that you see the object, you're viewing ...
Ok maybe I do see it. If one end of the object is separated by a distance x and the other end is separated by a distance $x+ \Delta x$, you'll be seeing both (using my amendment with the light source "being" the observer):
1)light that left the observer at $T'_{1}$, hit the base (close end) of the rod at $T'_{b}$ and hit the observer at $T'_{b \ arrives}$
2)light that left the observer at $T'_{2} = T'_{1} - \frac{2L}{c}$ hit the top (far end) at $T'_{top} = T'_{b} - \frac{L}{c}$ and hit the observer at $T'_{top \ arrives} = T'_{b \ arrives}$

So the length of our object's length at a given time would be ... Ok with these values our measuremts would be arbitrary.

We need to send out 2 signals simultaneously,

Light leaves at $T'_{1}$ (for both signals) hits the base at $T'_{b}$ and hits the top at $T'_{top} = T'_{b} + \frac{L'}{c}$ The first signal returns at $T'_{b \ arrives}$ and the second at $T'_{top \ arrives} = T'_{b \ arrives} + \frac{2L'}{c}$ and then $L' = c\frac{T'_{top \ arrives} - T'_{b \ arrives}}{2}$

The only thing I'm not sure of at this moment is that we'll have to take into account the distance traveled by our observer in the time between the first and second signal arriving. I'm just not sure how to do it without playing with it for a bit. However, these equations should hold for an observer at rest.

Hmmm... That's interesting. I wouldn't have guessed that particular example to be an example of 'length dilation', but I did guess it to be an effect of geometry, as I think I mentioned above.

17. Jun 6, 2014

### WannabeNewton

Length contraction is bounded below by zero and bounded above by one. It can be brought arbitrarily close to zero and of course it equals one in the rest frame of the object undergoing length contraction. Similar considerations hold for time dilation.

What you're talking about here, with regards to velocity in approaching or receding from light emanating from a ruler, is not space-time geometry but rather optics. If you move towards the ruler in a direction parallel to its length then it looks longer because of a simple optical effect; if you move perpendicularly to its length then it looks shorter again because of this optical effect. This is quite different from length contraction which is an effect of choosing a specific foliation of space-time. In particular, this is not peculiar to relativity. Optical effects modifying the appearance of objects, producing an apparent length, exist in Newtonian mechanics too.

18. Jun 6, 2014

### D H

Staff Emeritus
That is the heart of your problem. You are treating space and time as Newtonian. You cannot do that. The Newtonian description is only approximately valid, and then only when speeds are small compared to the speed of light. The Newtonian approximation falls apart when relative velocities are a significant fraction of the speed of light.

19. Jun 6, 2014

### BiGyElLoWhAt

Yes I am. I was trying to reach a similar conclusion with a classical approach. I'm not familiar enough with the curvature of Space-Time to be able to do any sort of calculations. So if an object is moving in a straight line in space and a straight line in time, it's actually a curved line through space-time, correct? But by how much is the real question. Is it necessary to take into account gravitational fields in order to solve this? I've read that space-time is altered while within a gravitational field.

To be honest I'm not really sure where to begin with this, but I'd really like to try to work through at least some of this without looking up the "answers".

Is space-time modelled as having some sort of "tensile strength"?

I guess can someone give me an idea of what I'm looking for without it being too big of a hint? Because so far what I've gathered is that I'm not going to get there with a classical approach (I can't say I'm all that surprised by this).

Or better yet, what kind of an observation is a good starting point in trying to recreate the concept of Space-time curvature and thus the effects that are caused by it?

20. Jun 6, 2014

### PAllen

No, in special relativity, straight is straight. The difference from what you are used to is the distance formula. You are thinking:

distance = √ (Δx2 + Δy2 +Δz2)

For spacetime, if you are talking about a length, you have (suppressing z):

proper length/distance = √ (Δx2 + Δy2 -c2Δt2)

For a time interval yo have:

proper time = √ (Δt2 - Δx2/c2 - Δy2/c2)

21. Jun 6, 2014

### stevendaryl

Staff Emeritus
I'm a little confused about what you're trying to do here. Are you trying to DERIVE the formulas for length contraction and time dilation, or are you just trying to understand what they mean?

What they mean is that a clock that runs at the normal rate as measured in its own reference frame runs slower by a factor of $\sqrt{1-(v/c)^2}$ as measured in a frame where the clock is moving at speed $v$. A stick whose length is one meter as measured in its own reference frame has length $\sqrt{1-(v/c)^2}$ as measured in a frame where the stick is moving at speed $v$ in a direction parallel to the direction the stick is pointing. Talking about light signals is one way to derive these results, but the results themselves don't have anything to do with light, particularly.

22. Jun 6, 2014

### BiGyElLoWhAt

If I'm not mistaken, this assumes that the curvature is of the x-y plane?

Is the negative sign a mistake? Or not?
If not this seems to imply that space time is always compressed, vs. simply warped (warped could be extended as well)

$\sqrt{ (\Delta x - \Delta t\ c\ \text{sin}(\Phi ) \text{cos}(\theta ) )^2 + (\Delta y - \Delta t\ c\ \text{sin}(\Phi ) \text{sin}(\theta ) )^2 + (\Delta z - \Delta t\ c\ \text{cos}(\Phi ) )^2}$

?

23. Jun 6, 2014

### BiGyElLoWhAt

I'm trying to derive the formula's without looking them up, but I'm basically trying to see if I can "logic" my way through this (conceptually), first, and then derive the mathematical models. Unfortunately I don't really have a good idea of where to start.

24. Jun 6, 2014

### mattt

No, Minkowski spacetime is a Lorentz manifold of zero curvature.

It is not a mistake. You need to look at the definition of a Pseudo-Riemannian Manifold. Space-times (in SR and also in GR) are a subset of Semi-Riemannian Manifolds, called Lorentz Manifolds. These can have zero curvature or non-zero curvature, depending of the concrete Lorentz Manifold you have. For example, in SR it is used a Minkowski space-time, which is a concrete Lorentz Manifold of zero curvature.

No.

You need to look at some mathematical concepts in Differential Geometry. "Semi-Riemannian Geometry with Applications to Relativity" (Barrett O'Neill) is great, for example.

You'll love it when you understand it all.

25. Jun 6, 2014

### pervect

Staff Emeritus
Relativity assumes that the speed of light is constant for all observers. I don't believe this is compatible with what you are assuming. I think you'll find that with your assumptions that if you have a system where motion is constrained to be in one spatial dimension, velocities always add. This can't be true in relativity, because we know that in relativity the speed of light is always a constant, and this won't be true if velocities always add.

[add]In relativity c + (any velocity <=c) = c

If you really want to learn relativity, you won't necessarily need "formal training", but you're actually going to need to actually read a book (like Bondi's, which is at high-school level as far as math goes) and pay attention to it.

Last edited: Jun 6, 2014