# Length contraction AP French

1. Apr 16, 2013

### Per Oni

Hi all,

Page 94 of Special relativity by AP French has a worked out example that puzzled me.
This is the example:

Frame S’ has a speed v=0.6c relative to S. Clocks are adjusted so that t=t’=0 at x=x’=0. Two events occur.
Event 1 occurs at x1=10m, t1=2E-7s (y1=0, z1=0).
Event 2 occurs at x2=50m, t2=3E-7s (y2=0, z2=0).

Question: What is the distance between the events as measured in S’?

It proceeds to work out Gamma=5/4.
Then using the formula given:
x2’-x1’=Gamma[(x2-x1) –v(t2-t1)] Therefore we get:
x2’-x1’=5/4[(50-10)-3/5(3 X 1E8)(3-2)1E-7]=27.5m

At first sight it looked ok to me because the length has been reduced from 40m to 27.5m. That is until I did the same example but now with t1=0 and t2=0. In that case the formula reads:
x2’-x1’=5/4[(50-10)]=50m.
In this case we get length expansion.

My take on this is that S’ sees S speeding away. Therefore S’ should measure a length contraction and not expansion for the case of t1=t2=0. Surely the events given are measured in S?

2. Apr 16, 2013

### Staff: Mentor

For S' to see simple length contraction, the events must be simultaneous in S'. In your example, the events are simultaneous in S. So S observes the length contraction.

3. Apr 16, 2013

### Per Oni

Hi Doc Al

I did realise before I asked this question that simultaneous events in S are not necessarily so in S’. I’ve got no problem with that.

These events happen in S. How can an observer in S see length contraction while x1 and x2 are stationary in S?

Furthermore, at which length does the formula used break down? If it works well for t1=2s and t2=3s, why not for t1=t2=0s?

4. Apr 16, 2013

### Staff: Mentor

Good.

Events don't happen in a frame. They just occur. They happen in all frames. Either frame, S or S', can describe the events. (That's what the Lorentz transformations do--translate one frame's descriptions to that of another frame.)
The events just happen to be simultaneous in S. They happen at x1 and x2. In particular, Δx = 40 m and Δt = 0.

Using the LT, you can calculate where and when those events happen according to S'. They take place at locations x1' and x2'. (Those locations are 'stationary' in S', just like the locations x1 and x2 are stationary in S.) It turns out that Δx' = 50 m but Δt' ≠ 0.

This makes a perfect illustration of the length contraction formula: S observes events to happen simultaneously at points that are 50 m apart according to S'; by simple length contraction L = γL0, so S must see the distance between those points as contracted to .8*50 = 40 m. It all works out fine.

In the previous case, it was certainly true that the distance between the events was shorter in S' than in S, but that's not a simple case of length contraction. (If it was, then the distance would have been .8*40 = 32 m, not 27.5 m.) In general, the transformation between S and S' is a combination of length contraction, time dilation, and the relativity of simultaneity, as described by the full Lorentz transformations. Only in special cases can you apply the simple length contraction or time dilation formulas by themselves.

5. Apr 17, 2013

### Per Oni

Yeah thanks for that. I have to be more careful with my lingo.

Not so. Those locations are stationary in S but not in S’.

Last edited by a moderator: Apr 17, 2013
6. Apr 17, 2013

### Staff: Mentor

I'm going to assume that you misread what I wrote. I'll rephrase:
The locations x1 and x2 are stationary in S. Similarly, the locations x1' and x2' are stationary in S'.

Where do the events happen? According to S, they happen at x1 and x2; According to S', they happen at x1' and x2'.

7. Apr 17, 2013

### ghwellsjr

I get the impression that you have not read the chapter before and after the example. Prior to the example, French spent a lot of time trying to disassociate the concept of an observer from a frame and to point out that the coordinates of an event are not anything that any observer can see.
This example is specifically not about length contraction. It is merely to show how coordinates are different in different frames and carry no additional meaning but are arbitrary. Read the text following the example, such as the last paragraph on page 95. The next page is where they get into Length (Lorentz) Contraction.

8. Apr 18, 2013

### Per Oni

I had not read this chapter, I dusted off this book to check up on something completely unrelated, and just leafed through when I saw that example.

All this while the example in question has very precisely related point events. All x,y,z,t in both frames are accurately given. Confusing?

@Doc Al:
I did misquote you, but I still cannot see how that solves the two questions I asked in post 3.

9. Apr 18, 2013

### Staff: Mentor

10. Apr 18, 2013

### ghwellsjr

Why do you say the pair of events are very precisely related when French just said they were unrelated? He just picked them out at random and he did this to show that when you transform their coordinates to another arbitrarily selected frame, they're still random.

Are you thinking since there is a precise algorithm to relate the coordinates of a single event in one frame to the coordinates of the same event in another frame, that that means the two sets of coordinates for the one event are very precisely related? Unfortunately, French didn't do this for a single event, instead he did it for the difference between a pair of events and I think maybe this led to the confusion.

In any case, even when a pair of events are related and defined in one frame of reference, we can pick any arbitrary frame moving with respect to the original one and get a new set of arbitrary coordinates. Notice his reference to "some arbitrary frame of reference" further down in the paragraph.

11. Apr 19, 2013

### Per Oni

The case you ascribe above is true if:
S’ sees a stationary object of 50 m and S’ is having a speed of .6c wrt S. Because then S will see that distance contracted to 40 m.

But the example given is the other way round. S’ sees an object having a velocity of .6c. Therefore S’ must see the length contraction.

12. Apr 19, 2013

### Per Oni

That bit further down the paragraph explains indeed what I’ve been missing so far. (I think).

If the two point events in the example (x1 and x2) are a measure of the length of a particular (solid) object and time interval (t1 and t2) are taken by one particular clock, in that case the point events are properly related?

If my take of this situation is true, do we need a different set of formulas to deal with the last case?

13. Apr 19, 2013

### Staff: Mentor

You have it backwards. According to your own calculations:
That means Δx' = 50 m. That's a "length" measured in S'. (S' does see a stationary object of 50 m!) Since Δt = 0, S measures the locations of those points at the same time in his frame. Thus S sees the length contraction.

14. Apr 19, 2013

### Staff: Mentor

The same Lorentz transformations are used with any pair of events.

However, as French notes, if you set things up so that x1 and x2 represent the length of an object or alternatively that t1 and t2 represent the ticks of a particular clock, then you will be able to show length contraction or time dilation. (Note that these are different set ups.)

In those special cases, the Lorentz transformations simplify to become the "length contraction" and "time dilation" formulas.

15. Apr 19, 2013

### ghwellsjr

Good.
In order for two events (we can drop the word "point" to fall in line with common usage) to be a measure of the length of an object, the two time coordinates have to be the same. Similarly, in order for them to be a measure of a time interval by a clock, the spatial coordinates must be the same. Clearly, both cannot be true for the same two events.

It turns out that for any pair of random events, we can calculate the Spacetime Interval between them and it will be one of three cases: space like, time like, or light like. For space like, there is a frame in which the two events can be the measure of the length of a stationary object. For time like, there is a frame in which the two events can be the measure of a time interval on an inertial clock. For light like, neither case applies and the two events define the path of a photon. Why don't you see which case the example applies to?
No, you just need to find the specific speed that you need to transform the original frame into to make it apply to a length or a time interval. Do you think you can do that?

I would suggest that you convert the coordinates of the example into feet and nanoseconds and define the speed of light to be 1 foot per nanosecond to make the computation easier. Then the original events will have coordinates of 32.8 feet @ 200 nS and 164 feet @ 300 nS.

Good luck.