Does Time Dilation Affect Distance Measurements in Special Relativity?

[aintnuthin]

Of course he 'acknowledges' that he and Harriet are in relative motion. But that's got nothing to do with his measurement of the distance between the clocks, since the clocks are at rest with respect to him.

I agree, it doesn't affect the measurements that I make in my frame in the least. But it could affect how I attempt to impute (transform) those measurements to her frame via calculation, couldn't it?

 

[Doc Al]

Well, I guess I'm not sure what is unclear here. My logic is kinda like this.

1. If any two objects are moving with respect to each other, at least one of them must be "really" moving, even if both are always "at rest" with respect to themselves.

Nope. The concept of "really" moving is meaningless. There is only relative motion.

2. Maybe you know who's really moving, maybe you don't, but, either way, you know one of the two is.

Nope.

3. If you don't "know" then let's say the odds are 50-50 that you are moving with respect to the other, and that it is "stationary" with respect to you.

Nope.

Everyday example: I am going down the freeway at 60 mph. As between me and the road, I know ONE of us is (relatively) stationary and that one of us is (relatively) moving. In that case, I would assume that I'm the one moving relative to the road (and all other fixed things on the planet that I see), not that they are all coming to, and going past, me. In such a case I would NOT employ the LT from the perspective that *I* was stationary and that something fixed (say a tree) that was "approaching" me was "really moving."

Nope. An observer always measures things from his own frame.

On the contrary, I would employ my knowledge of SR (such as it is) to calculate times and lengths from "in reverse." I would assume that Earth's clocks would be faster (not slower) than mine and that it's lengths would be longer (not shorter) than mine are in my frame. That, for example, there would be some (however infintesimal) difference in the rate at which my watch keeps time, and the rate it keeps time when I come to a stop sign. As I sped up, I would assume that my watch was running slower than the one I saw the guy standing at the stop sign wearing. I would assume that his watch was now faster than mine, and mine slower than his. Would I be wrong to make such assumptions?

Yep. All wrong.

 

[Doc Al]

I agree, it doesn't affect the measurements that I make in my frame in the least. But it could affect how I attempt to impute (transform) those measurements to her frame via calculation, couldn't it?

All you need to know is the relative speed between the two frames.

 

[Dale]

Well, I guess I'm not sure what is unclear here. My logic is kinda like this.

1. If any two objects are moving with respect to each other, at least one of them must be "really" moving, even if both are always "at rest" with respect to themselves.

This is wrong, you must get rid of this idea. It completely misses the point of relativity, not just Einstein's special relativity but also Galilean relativity of classical physics.

Even in Newtonian physics there is no such thing as "really" moving, only moving relative to something else. This is called the "first postulate" or the "principle of relativity" and has been experimentally verified fact since Galileo's time:
http://en.wikipedia.org/wiki/Principle_of_relativity

It seems that all of your errors stem from this point. This is the fundamental symmetry I was describing above. If you want to learn relativity you definitely need to learn this point well.

 

[aintnuthin]

Nope. The concept of "really" moving is meaningless. There is only relative motion.


Nope.


Nope.


Nope. An observer always measures things from his own frame.

As i said (next), I agree that an observer always measures things from his own frame. But, likewise, he always imputes (to others) from his own frame, too. How, and what, he imputes would seem to depend on his assumptions.

Doc, that's sure a lot of "nopes," without any elaboration on your part. Is what I'm suggesting impossible?

 

[Doc Al]

As i said (next), I agree that an observer always measures things from his own frame. But, likewise, he always imputes (to others) from his own frame, too. How, and what, he imputes would seem to depend on his assumptions.

The only 'assumption' is the relative motion between the frames.

Doc, that's sure a lot of "nopes," without any elaboration on your part. Is what I'm suggesting impossible?

Yes. You are mixing up measurements in different frames. The key problem is your assumption that one of the frames is 'really' at rest. See DaleSpam's response.

If your initial statement is wrong, and your following statements depend on the first, then there's not much to elaborate on.

 

[aintnuthin]

This is wrong, you must get rid of this idea. It completely misses the point of relativity, not just Einstein's special relativity but also Galilean relativity of classical physics.

Even in Newtonian physics there is no such thing as "really" moving, only moving relative to something else. This is called the "first postulate" or the "principle of relativity" and has been experimentally verified fact since Galileo's time:
http://en.wikipedia.org/wiki/Principle_of_relativity

It seems that all of your errors stem from this point. This is the fundamental symmetry I was describing above. If you want to learn relativity you definitely need to learn this point well.

OK, Dale, your assertions are noted. I don't agree with them all, but don't wish to get sidetracked on those topics. Galileo's "principle of relativity" was a mere by-product of his principle of inertia, which he used (effectively) in support of his argument that the Earth was "really" revolving around the sun, for example. But again, I just trying to get answers to a very narrow, limited question right now.

 

[Doc Al]

I don't agree with them all, but don't wish to get sidetracked on those topics.

Sidetracked?? Until you disabuse yourself of the notion of something 'really' being at rest, don't expect to make much progress.

 

[aintnuthin]

The only 'assumption' is the relative motion between the frames.

Doc, this assertion that there is only one assumption has confused me from the outset. After asserting that, initially, you proceded to say Harriet's measurements were determined "because she was stationary." How could she (or you) know that?

 

[Doc Al]

Doc, this assertion that there is only one assumption has confused me from the outset. After asserting that, initially, you proceded to say Harriet's measurements were determined "because she was stationary."

Where exactly did I say that? All one can ever say is that something is or isn't stationary with respect to something else. (Of course Harriet, and everyone else, treats themselves as being stationary in their own frame.)

Are you back to being confused about the terminology of 'stationary' frame versus 'moving' frame? These are just relative terms. To Henry, he's the 'stationary' frame; to Harriet, she is. No one is talking about being 'really' stationary--that's meaningless.

 

[One Brow]

When I first posted, aintnuthin pretty much assured me you would correct my erroneous notions. Thank you all for the explanations offered. I was not aware he planned on following me here. I won't trouble you again.

 

[ghwellsjr]

This conversation has become boring so I will work the problem.
Worldline of clock 1
$$x'=0$$
$$1.25 (0.6 c t + x) = 0$$
$$x=-0.6 c t$$

Worldline of clock 2
$$x'=6$$
$$1.25 (0.6 c t + x) = 6$$
$$x = 4.8 - 0.6 c t$$

So we immediately see that the distance between the clocks is 4.8 lightseconds.

Worldline of Harriet
$$x' = 0.6 c t$$
$$1.25 (0.6 c t+x)=0.75 c \left(\frac{0.6 x}{c}+t\right)$$
$$x = 0$$

Finding the intersection of Harriet's worldline with clock 1 we find:
$$0 = -0.6 c t$$
$$t = 0$$

Finding the intersection of Harriet's worldline with clock 2 we find:
$$0 = 4.8 - 0.6 c t$$
$$t = 8/c$$

So we see that the time was 8 seconds.

Claimant B is correct. QED.

Your analysis is so simple. I used the Lorentz Transform to analyze the same scenario but as you can see it is very much more complicated although I do get the same answer. Could you provide insight into how you do it so much more easily?

Here's how I explain the process in detail:


A Frame of Reference is a set of x, y, z coordinates along with time that we use in Special Relativity to define and analyze a scenario. Each set of [t, x, y, z] coordinates is called an "Event". All observers and objects (clocks & rulers) exist in any particular FoR that we are considering.

When someone refers to Harry's FoR, they are not excluding Harriet, they are merely saying that Harry is a rest in that particular FoR and Harriet is moving. In order to talk meaningfully about Harry's FoR, we need to specify each significant event for Harry and each significant event for Harriet, all within the same FoR.

When people then talk about Harriet's FoR, they are talking about a totally different FoR in which Harriet is stationary and Harry is moving. The coordinates in these two frames are different and you use the Lorentz Transform to convert all the events from one FoR to all the same events in the other FoR.

For simplicity's sake, for scenarios like the yours, since we can assign everything to the x coordinate, we can ignore y and z components (because we can set them everywhere to zero).

So let's go back to the first post in this thread and compile the events defined there according to Henry's FoR (remember, these coordinates are for t in seconds and x in light-seconds as [t,x]):

[0,0] Location of clock1 at start of scenario
[0,6] Location of Henry and clock2 at start of scenario
[10,0] Location of clock1 at end of scenario
[10,6] Location of Henry and clock2 at end of scenario
[0,0] Location of Harriet and her clock at start of scenario
[10,6] Location of Harriet and her clock at end of scenario

Note that the length of time for the scenario to progress in Henry's FoR is the distance between Henry's two clocks divided by the relative speed which is 6/0.6 or 10 seconds.

So now we use the Lorentz Transform on each of these six events in Henry's FoR to see what they are in Harriet's FoR.

First we want to calculate gamma, γ, which is equal to 1/√(1-ß²) where ß is the speed as a fraction of c. So:

γ = 1/√(1-ß²) = 1/√(1-0.6²) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

Then we use these two formulas for each t and x coordinate:

t' = γ(t-vx/c²)

x' = γ(x-vt)

But since we are using units in which c=1 we can simplify them to:

t' = γ(t-vx)

x' = γ(x-vt)

Note that t' and x' are the new values based on the old values, γ, t, x, and v.

The velocity, v, is 0.6 because we want Harriet to move from an x value of 0 at the start of the scenario to an x value of 6 at the end of the scenario (according to Harry's FoR).

Now we do the detailed calculations. Some of the events are used twice so we only have to do four sets of calculations.

[0,0]:
t=0 and x=0
t' = γ(t-vx) = 1.25(0-0.6*0) = 0
x' = γ(x-vt) = 1.25(0-0.6*0) = 0

[0,6]:
t=0 and x=6
t' = γ(t-vx) = 1.25(0-0.6*6) = 1.25(-3.6) = -4.5
x' = γ(x-vt) = 1.25(6-0.6*0) = 7.5

[10,0]:
t=10 and x=0
t' = γ(t-vx) = 1.25(10-0.6*0) = 12.5
x' = γ(x-vt) = 1.25(0-0.6*10) = -7.5

[10,6]:
t=10 and x=6
t' = γ(t-vx) = 1.25(10-0.6*6) = 1.25(10-3.6) = 1.25(6.4) = 8
x' = γ(x-vt) = 1.25(6-0.6*10) = 0

To summarize the four sets of calculations as [Henry's FoR] > [Harriet's FoR]:
[0,0] > [0.0]
[0,6] > [-4.5,7.5]
[10,0] > [12.5,-7.5]
[10,6] > [8,0]

And substituting the coordinates from Henry's FoR to Harriet's FoR:
[0,0] Location of clock1 at start of scenario
[-4.5,7.5] Location of Henry and clock2 at start of scenario
[12.5,-7.5] Location of clock1 at end of scenario
[8,0] Location of Henry and clock2 at end of scenario
[0,0] Location of Harriet and her clock at start of scenario
[8,0] Location of Harriet and her clock at end of scenario

From this you can see that the time from start to end of the scenario for Harriet is 8 seconds.

There are many ways that Harriet can measure the distance between Henry's two clocks. Probably the simplest is for her to measure how long it takes between clock1 and clock2 arriving at her location and knowing the speed of the clocks (and Henry) she can calculate the distance as simply the speed multiplied by the time interval. This would be 0.6 times 8 or 4.8 light-seconds.

These values agree with the ones of Claimant B from post #1.

However, you may be wondering why the values for Claimant A aren't correct because we do see a location value of 7.5 light-seconds and a time of 12.5 seconds in Harriet's FoR. Well the reason why those values are not legitimate answers is because they are for events that happened simultaneously in Henry's FoR but they are not simultaneous in Harriet's FoR. For the distance between clock1 and clock2 we have to calculate a pair of events for those two clocks that occurred at the same time in Harriet's FoR. We can pick any time we want but it makes sense to use the first event where the time is zero and the location of clock1 is 0. Then all we have to do is figure out where clock2 is at time zero. We use a normal interpolation technique to do this. Looking at the second and fourth events, we find clock2 starts out at a location of 7.5 at a time of -4.5 and ends up at location 0 at a time of 8 seconds. The total time is 8 - (-4.5) = 12.5 and the total distance is 7.5 light-seconds. We want to know the ratio of 4.5 seconds out of 12.5 seconds, which is 0.36. Now if we multiple this ratio times the 7.5 light-seconds we get 2.7 light-seconds distance from the starting location which is 7.5 so 7.5 minus 2.7 or 4.8 light-seconds is the location of clock2 at time zero and that is the answer we are looking for.

 

[Dale]

Your analysis is so simple. I used the Lorentz Transform to analyze the same scenario but as you can see it is very much more complicated although I do get the same answer. Could you provide insight into how you do it so much more easily?

I am glad that you liked it!

I looked through your analysis and it is correct also, there is not really any difference between a four-vector analysis like yours and an algebraic analysis like mine. The four-vectors are just a compact way to write the algebra. The only reason I used the algebraic approach is that I figured that aintnuthin could follow algebra, but would probably get confused by four-vectors.

So now we use the Lorentz Transform on each of these six events in Henry's FoR to see what they are in Harriet's FoR.

First we want to calculate gamma, γ, which is equal to 1/√(1-ß²) where ß is the speed as a fraction of c. So:

γ = 1/√(1-ß²) = 1/√(1-0.6²) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

Then we use these two formulas for each t and x coordinate:

t' = γ(t-vx/c²)

x' = γ(x-vt)

One tip for making it more concise. If you are using four-vectors then you can cast the Lorentz transforms as matrices instead of as equations.

Now we do the detailed calculations. Some of the events are used twice so we only have to do four sets of calculations.

[0,0]:
t=0 and x=0
t' = γ(t-vx) = 1.25(0-0.6*0) = 0
x' = γ(x-vt) = 1.25(0-0.6*0) = 0

[0,6]:
t=0 and x=6
t' = γ(t-vx) = 1.25(0-0.6*6) = 1.25(-3.6) = -4.5
x' = γ(x-vt) = 1.25(6-0.6*0) = 7.5

[10,0]:
t=10 and x=0
t' = γ(t-vx) = 1.25(10-0.6*0) = 12.5
x' = γ(x-vt) = 1.25(0-0.6*10) = -7.5

[10,6]:
t=10 and x=6
t' = γ(t-vx) = 1.25(10-0.6*6) = 1.25(10-3.6) = 1.25(6.4) = 8
x' = γ(x-vt) = 1.25(6-0.6*10) = 0

That makes all of these into matrix multiplications.

And substituting the coordinates from Henry's FoR to Harriet's FoR:
[0,0] Location of clock1 at start of scenario
[-4.5,7.5] Location of Henry and clock2 at start of scenario
[12.5,-7.5] Location of clock1 at end of scenario
[8,0] Location of Henry and clock2 at end of scenario
[0,0] Location of Harriet and her clock at start of scenario
[8,0] Location of Harriet and her clock at end of scenario

From this you can see that the time from start to end of the scenario for Harriet is 8 seconds.

For the time dilation your approach is nice and easy.

There are many ways that Harriet can measure the distance between Henry's two clocks. Probably the simplest is for her to measure how long it takes between clock1 and clock2 arriving at her location and knowing the speed of the clocks (and Henry) she can calculate the distance as simply the speed multiplied by the time interval. This would be 0.6 times 8 or 4.8 light-seconds.

This is probably the most direct way to get the distance from what is given, but it is still somewhat indirect. The problem is, of course, the relativity of simultaneity. Length is the distance between two worldlines, so when I do a problem that involves length contraction I like to use worldlines instead of events. You can still do that using the four-vector notation:

Worldline of clock 1:
[t',0] -> [t,-0.6t]

Worldline of clock 2:
[t',6] -> [t,4.8-0.6t]

Worldline of Harriet
[t',0.6t']->[t,0]

Where each [] is now the parametric equation of a line rather than a single event. The one disadvantage to this approach is that it hides the relativity of simultaneity stuff that the event-wise approach highlights.

 

[Dale]

When I first posted, aintnuthin pretty much assured me you would correct my erroneous notions. Thank you all for the explanations offered. I was not aware he planned on following me here. I won't trouble you again.

No worries, it is hard for someone who doesn't understand the principle of relativity to understand the rest of relativity.