# Question on time measured

1. Aug 1, 2011

### One Brow

On my blog, I am having a continuing discussion on special relativity with another person (we are both amatuers). We disagree over what Jill measures below.

Scenario: Harrietl is stationary, with a clock. Henry and clock1 are moving at the same velocity, whose speed is .6c with respect to Harriet. Henry measures the distance between Henry and clock1 as 6 light-seconds. Henry has a clock (clock2) synchronized with clock1. As clock1 passes Harriet, Harriet synchronizes her clock to clock1 (and for the sake of the discussion, let's call that time 0). Then, Henry passes Harriet.

Claimant A: Harriet measures the distance between Henry and clock1 as 7.5 light-seconds and the time it takes them to pass as 12.5 seconds.

Claimant B: Harriet measures the distance between Henry and clock1 as 4.8 light-seconds and the time it takes them to pass as 8 seconds.

2. Aug 1, 2011

Not correct.

Correct.

3. Aug 1, 2011

### aintnuthin

Doc Al. Thanks for your answer. Can you elaborate at all on the reasoning and methods you employed to reach your conclusions.

I am "claimant A" in this disagreement, and do not understand the grounds for your response.

Can anyone explain this?

4. Aug 1, 2011

### Staff: Mentor

It's just basic relativity. For the distance, use length contraction. For the travel time, use distance = speed * time.

Why don't you explain your reasoning.

5. Aug 1, 2011

### aintnuthin

I could probably phrase my reasoning in a number of different ways, but let me ask this first.

It is my understanding that, according to the lorentz transform, the "stationary" party will see the moving party to have contracted lengths and dilated (slowed) time.

Is that correct?

If so, you have at least implicitly made Henry the stationary party, right? His measurement of time is greater (10 seconds versus 8) and his measurement of distance is greater (6 LS vs 4.84 LS).

How have you determined that Henry is (relative to Harriet, at least) stationary and that, therefore, Harriet is the moving party?

6. Aug 1, 2011

### Staff: Mentor

Yes.

No. "Stationary" and "moving" are relative terms. Either one may consider themselves as the 'stationary' party. Since Harriet is doing the measuring, she is the stationary party. As far as she is concerned, Henry is the one who is moving.

Harriet is measuring the distance between points in Henry's frame; since Henry is moving with respect to her, she measures a shorter distance between those points.

You don't need to worry about time dilation for this problem.

7. Aug 1, 2011

### aintnuthin

First of all, let me clarify the question as I see it. I agree that she can "consider" herself to be "moving," but my question is not about subjective "considerations" by either party. It is basically about the validity of the lorentz transform, and it's pertinence, if any, when attempting to solve such problems.

First question: You say:"... she is the stationary party." If she is the stationary party, why are her time and distance both less than Henry's? Wouldn't that be contrary to the Lorentz transform?

Second question: If Henry "considered himself" to be stationary, what would Harriet's time and distance be from that perspective. My answer to that would be the one you gave, i.e., 8 seconds and 4.84 LS for Harriet.

Third question: You say: "Harriet is measuring the distance between points in Henry's frame." But Henry is also "measuring the distance," isn't he. In fact, as I read the question, prior to encountering Harriet, he has set up a clock 6 LS away from him in his frame, and he has synchronized a second clock, which he keeps with him. Are his "measurements" somehow invalid?

Thanks again for your help and participation, Doc Al.

8. Aug 1, 2011

### aintnuthin

One more thing: Implicit in all of this was the assumption that the time on Henry's clocks would be 10 seconds elapsed. Perhaps that wasn't explicitly stated. This conclusion would be based on the following reasoning:

Moving or not, the clocks are 6 light seconds apart in Henry's frame. Therefore it is basically tautological that HE would measure the time elaspsed in the duration to be 10 seconds. (6/.6 = 10)

9. Aug 1, 2011

### Staff: Mentor

If you don't like the terms 'stationary' and 'moving', then use 'primed' and 'unprimed'.

In any case, the Lorentz transformations allow us to transform measurements made in one frame to the corresponding measurements in another frame that is moving with respect to the first. In this case we are given the length measurement made by Henry in his frame; we can use the 'length contraction' formula to get the length as measured by Harriet. (Note: The length contraction formula is a special case of the Lorentz transformations.)

She's measuring the length of a moving system, so it is length contracted.

Henry does consider himself to be stationary. (And so does Harriet!)
That's the only answer there is as to what Harriet measures.

Henry's measurements are perfectly valid. In fact we used them to figure out what Harriet would measurement. The only measurement given is the distance measured by Henry.

10. Aug 1, 2011

### Staff: Mentor

Right. According to Henry, the travel time for Harriet to move from one clock to another will be 10 seconds. But realize that according to Harriet, Henry's clocks are not synchronized.

You cannot simply apply the time dilation formula to figure out what Harriet will measure for the travel time without taking clock synchronization into account. (Of course, since we have the speed and the distance, we don't need to apply time dilation at all to figure out Harriet's travel time measurement.)

11. Aug 1, 2011

### aintnuthin

This where you completely lose me. If she is stationary, and if, with respect to her frame, Henry's length is contracted, and if, in her frame the distance is 4.8 light seconds, why doesn't she calculate his length to be 3.87 light seconds.

You say: "In this case we are given the length measurement made by Henry in his frame; we can use the 'length contraction' formula to get the length as measured by Harriet. " I totally agree with this. It's just that, if she is "really" stationary, then her length should be LONGER, not shorter than his, as I see it. If his distance is 6, hers should be 7.5. That way, when she uses the LT to "calculate" the length in Harry's frame, it will come out to be 6 light seconds, just as he in fact measures it to me.

See what I'm getting at?

12. Aug 1, 2011

### aintnuthin

Yes, I think I realize that, according to Harriet Henry's clocks are not synchronized. But the reverse is just as true, isn't it? Harriet's clock is not synchronized to his. When the two meet, if Henry see's Harriet's clock to read 12.5, while his reads only 10, he will insist that her clock must have started at 4.5 seconds, not 0. 12.5 - 4.5 = 8 (the time he believes must have passed in her frame IF he considers himself to be stationary.

13. Aug 1, 2011

### Staff: Mentor

We are given the length as measured by Henry. So, via length contraction, Harriet measures that length be shorter. Note that length measurements are essentially measurements of the positions of two points at the same time. Henry says that Harriet's measurements were not taken at the same time--so he cannot simply apply the length contraction formula to Harriet's measurement. (It's the clock synchronization issue again.)

14. Aug 1, 2011

### aintnuthin

You say:"Henry's measurements are perfectly valid. In fact we used them to figure out what Harriet would measurement. The only measurement given is the distance measured by Henry."

Again, I agree completely with this statement. I am referring to it solely for the purpose of phrasing my question (or, stating my case, if you will) in a slightly different way, not because I disagree with this statement.

By hypothesis, Henry's time, in his frame is 10 seconds. By hypothesis, Henry's distance, in his frame, is 6 light seconds.

By hypothesis, Harriet is stationary and Henry is moving at .6c.

By virtue of the Lorentz Tranform, Harriet would have to see both Henry's time and his distance to be LESS (not more) than what her own measurements indicate.

Therefore, she cannot, per the LT, make measurements in her own frame which indicate that she, not Jack, has the shorter distance and slower time.

Therefore she will measure the time and distance in her frame to be 12.5 seconds and 7.5 light seconds, respectively.

Where, if anywhere, have I gone wrong in this train of thought?

Last edited: Aug 1, 2011
15. Aug 1, 2011

### Staff: Mentor

Try actually doing a Lorentz transform.

In the primed (Henry's) frame the path (worldline) of clock 1 is:
x'=0

The worldline of clock 2/Henry is:
x'=6

The worldline of Harriet is:
x'=.6ct'

Simply make the substitutions here and simplify in order to get the paths in the unprimed frame:
http://en.wikipedia.org/wiki/Lorent...ormation_for_frames_in_standard_configuration

16. Aug 1, 2011

### aintnuthin

Thanks for the suggestion, Dale, but I'm not a mathematician. It seems to me that x and x' prime can be virtually interchangable. Can you put your point into words, or can it only be understood mathematically?

I have heard many prominent physicists say that, in special relativity, the moving clock runs slow, and the stationary observer see the moving observer's lengths as being contracted. In this example, by hypothesis, Henry is the moving party.

So why doesn't Harriet see his time as slowed and his lengths contracted if she is stationary?

17. Aug 1, 2011

### Staff: Mentor

OK.
They are in relative motion; their relative velocity is 0.6c.

This is where you are messing up. You are confusing general time intervals and distances between space-time events, which must be handled via the full LT, with simple cases of moving lengths and clocks that can be handled via the simple time dilation and length contraction formulas. (Look up the full LT: they are more complicated than simply length contraction and time dilation.)

In this case, consider these two events:
(1) Harriet passes clock A (at rest in Henry's frame)
(2) Harriet passes clock B (where Henry is)

We know the distance between these events in Henry's frame: 6 light seconds.
We know the time between those events in Henry's frame: 10 seconds.

You can plug those values into the LT and compute Harriet's measurements for those two events:
She says the distance between those events is 0 (they occur at the same place in her frame)
She says the time between those events is 8 seconds.

If we want to know what Harriet would measure as the distance between clocks A and B, that's a different problem. We can use the LT for that as well, or take the short cut since it's a length that is stationary in Henry's frame and use the length contraction formula.

18. Aug 1, 2011

### Staff: Mentor

You don't have to be a mathematician, this is just middle-school level algebra. This excuse is not acceptable.

19. Aug 1, 2011

### Staff: Mentor

So far, so good.
Whether Henry is moving or not depends on who is the observer. Obviously to Harriet, Henry is moving.
She does! She measures the length between the clocks to be contracted. She also measures Henry's clocks as running slow.

20. Aug 1, 2011

### aintnuthin

Again, I am not a mathematician, but since you have brought up "invariant" spacetime intervals, I will present my understanding of them.

To begin with, you say:

"In this case, consider these two events:
(1) Harriet passes clock A (at rest in Henry's frame)
(2) Harriet passes clock B (where Henry is)"

By hypothesis, Harriet is not "passing" clock A. She is "being passed" by it, and I think there is a meaningful distinction, for spacetime interval calculations.

In that context, I would see it this way.

1. Jack's 2 clocks are not moving with respecct to him. Therefore, for the purposes of constructing invariant spacetime intervals, his distance (6 light seconds) will be considered to be the "proper length."

2. Jill's clock will be present at both events 1 and 2 and will therefore (again, for the purpose of constructing an invariant spacetime interval) be considered to have the "proper time."

3. You can only construct an invariant spacetime interval by taking the "proper length" from one frame Henry's, here), and the "proper time" from another (Harriet's, here).

To get an invariant interval, the "proper time" (from Jill's frame) would therefore have to be 12.5 seconds.

Is this wrong?