Does Time Dilation Affect Distance Measurements in Special Relativity?

[Dale]

By hypothesis, he is NOT stationary. He is moving, while Harriet is stationary. It makes a difference, doesn't it?

Not a bit. That is kind of the whole point of relativity. Labeling one frame "stationary" does not confer any special status on it, it is just an arbitrary label like "A" or "primed".

 

[aintnuthin]

Not a bit. That is kind of the whole point of relativity. Labeling one frame "stationary" does not confer any special status on it, it is just an arbitrary label like "A" or "primed".

It may not confer any ontological status upon it, but it certainly conveys an analytical status. Once you stipulate that a party is moving, then, for the sake of consistently analyzing the mathematical implications of the LT, etc., then you must maintain that stipulation, intact, throughout the analysis. You can't just start changing your assumptions, willy-nilly, during the course of analysis. To do so only invites all kinds of unwarranted confusion. Actually, it does more than merely "invite" confusion. It creates it, ipso facto.

 

[Dale]

It may not confer any ontological status upon it, but it certainly conveys an analytical status.

No it doesn't. All it does is determine the sign of v in the Lorentz transform equation that I linked to earlier. As long as you keep that straight there is no other ontological or analytical difference between the frames. All of the features such as time dilation, length contraction, and relativity of simultaneity work the same either way.

Spend the 20 minutes to do the math, it will be well worth your effort.

 

[aintnuthin]

No it doesn't. All it does is determine the sign of v in the Lorentz transform equation that I linked to earlier. As long as you keep that straight there is no other ontological or analytical difference between the frames. All of the features such as time dilation, length contraction, and relativity of simultaneity work the same either way.

Spend the 20 minutes to do the math, it will be well worth your effort.

Math depends on assumptions. Right now, I'm more interested in identifying the assumptions than I am in accepting them, without question, and then "doing the math."

Can I take your response to mean you believe that, for purposes of analysis (mathematical or logical) it is fine to change the underlying assumptions, at will, during the course of the analysis?

 

[Dale]

Can I take your response to mean you believe that, for purposes of analysis (mathematical or logical) it is fine to change the underlying assumptions, at will, during the course of the analysis?

When the assumption is a specific choice of a symmetry of the equations, then yes. In fact, this technique can be incredibly useful for solving otherwise difficult problems.

For example, if you have an equilateral triangle and you are calculating the area using 1/2 bh then (because it is symmetric) it doesn't matter which side you call the base. You can even use one side as the base to measure b then switch to another side to draw your perpendicular to find h. This is one of the reasons that symmetry is so important in math and physics.

 

[aintnuthin]

When the assumption is a specific choice of a symmetry of the equations, then yes. In fact, this technique can be incredibly useful for solving otherwise difficult problems.

For example, if you have an equilateral triangle and you are calculating the area using 1/2 bh then (because it is symmetric) it doesn't matter which side you call the base. You can even use one side as the base to measure b then switch to another side to draw your perpendicular to find h. This is one of the reasons that symmetry is so important in math and physics.

It's fine to recognize symmetry, when it exists. It is wrong to impute it by fiat when it does not exist.

Take this hypothetical "math problem" as an example:

John has 10 dollars in his bank account. After a $5 transaction, what is his new balance?

Standing alone, you can't answer that. As an analogy, that is like merely saying that two objects are "in relative motion." That's not enough information, and you need more than that. In this case, you will ask if the "transaction" was a deposit or a withdrawal before trying to answer.

So, I say "withdrawal."

Now suppose, for the purposes of solving this problem, you decide change the assumptions of the problem and treat the transaction as if were a deposit, rather than a withdrawal, as was stipulated. You can do this on the basis of absolute "symmetry," i.e., identity (5 = 5).

Your answer will then be "$15."

But your answer will be incorrect because it has changed the assumption which the question was based on. That not the "fair" way to do it. You insisting that $5 is $5, after all, will not buttress your case that the answer of $15 is correct

 

[Dale]

It's fine to recognize symmetry, when it exists. It is wrong to impute it by fiat when it does not exist.

On the other hand, take this hypothetical "math problem."

John has 10 dollars in his bank account. ...

Bank account balances are not symmetric under deposit-withdrawal parity. They are symmetric under transformations of 1 dollar into 100 pennies. You can do that symmetry operation as often as you like, even in the midst of a single transaction.

Physics is symmetric under boosts. That is the whole point of relativity.

 

[aintnuthin]

Bank account balances are not symmetric under deposit-withdrawal inversions.

Physics is symmetric under boosts. That is the whole point of relativity.

Your assertion of "symmetry" in this case leads to the erroneous conclusion that the lorentz transform, properly applied, will result in the stationary party imputing longer lengths and faster times to a frame of reference which is moving "relative" to it.

That is inconsistent with the generally accepted consequences of the LT, as it is standardly presented.

 

[Dale]

Says someone who has never actually done one.

I am sorry, but you are simply misunderstanding the LT and the standard presentations. It is easy to do, which is why working problems is valuable.

 

[aintnuthin]

Bank account balances are not symmetric under deposit-withdrawal parity.

My response would be that there is no "symmetry" between a reference frame which is presumed to be traveling at a speed of 0 and a frame which is presumed to be traveling at a velocity of .6c, either.

In the original problem, where it was stipulated that Henry was moving, the distance in his frame was 6 light seconds (implying a time of 10 seconds elapsed between events 1 and 2 in his frame).

THAT will not change, if you change your assumptions about his speed. Whether is is going 0 or .6c, or .9999c for that matter, he has still measured 6 light seconds distance and 10 seconds clock time in that frame. In terms of the analogy, it is the "$5 transaction."

But Harriet's time and distance will change, if you reverse the assumptions about who is moving.

Had the problem stipulated that Harry was stationary (rather than moving) then the correct answer with respect to Harriet's time and distance would indeed be 4.84 light seconds distance (and 8 seconds clock time). Call that a "withdrawal."

But the problem did NOTstipulate that Harry was stationary (but rather stated the opposite, i.e., that he was moving). By analogy, this would result in a "deposit" not a "withdrawal."

The answer (for Harriet--who might be viewed as the "bank balance" in this analogy) will therefore be different.

 

[aintnuthin]

I am sorry, but you are simply misunderstanding the LT and the standard presentations. It

Well, Doc Al seems to agree that I have the correct understanding:

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?

Yes.

Anybody else have any input to offer on this question? Doc Al seems to have headed out.

 

[Dale]

My response would be that there is no "symmetry" between a reference frame which is presumed to be traveling at a speed of 0 and a frame which is presumed to be traveling at a velocity of .6c, either.

Nature disagrees with you:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

But Harriet's time and distance will change, if you reverse the assumptions about who is moving.

Can you prove your assertion using the full Lorentz transform I linked to earlier?

 

[aintnuthin]

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

Doc, you say this is simple, but I can't see why. It seems abiguous to me, and it seems to me that this "simple" approach it is leading you to make inconsistent statements, which imply an infinite regression.

You say "use length contraction," but the unanswered question is "use it how?" Let's assume that you mean something like: "Whenever you see a length stated, in any context, then impute a lesser distance to all other frames." What results do we get?

Let's use a case where the relative speed is such (.866c, or whatever it is) that it results in a reduction of time and length of 50%. How does this work? Whose distance and time get reduced? Do the circumstances in which you came across the revelation of "length" matter at all?

Example 1: I say "a rocket is traveling at the rate of .866c (relative to Earth observers) to a star which is 8 light years away according to Earth observers"

So, you say: "stop right there. I've heard all I need to hear. That means the distance for the rocket is 4 light years."

Example 2: I say "a rocket is traveling at the rate of .866c (relative to Earth observers) to a star which it deems to be 4 light years away."

So, you say: "stop right there. I've heard all I need to hear. That means the distance for Earth observers is 2 light years."

Which is it? Do you ALWAYS just take whatever length you hear first, and then impute half that distance to the other party?

 

[aintnuthin]

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.

Doc, in light of your agreement that the stationary party measures the other (moving) party's distance to be shorter, when you say "she measures a shorter distance between those points" I take you to mean that, when she "calculates" (not "measures," strictly speaking) those distances, they will be shorter than what she measures them to be (in her frame). In order for that to be the case, the distance, as measured in her frame, would have to be LONGER than 6 LS. If she measured the distance, in her frame, to be 7.5 seconds, THEN 6 LS would be 80% of hers (i.e., "shorter"). As I read you, you say the same thing (in a slightly different way) here:

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

That is, it is contracted with respect to HER, not with respect to itself, right? As you correctly note elsewhere:

The only measurement given is the distance measured by Henry.

Yes, and that measurment is 6 light seconds. If she is stationary, she must measure a longer distance in her frame in order for 6 light seconds to be the lesser distance, right? That seems to be what you are saying here:

We are given the length as measured by Henry. So, via length contraction, Harriet measures that length be shorter.

The question is "shorter than what?" Again, it seems to me that, per the LT, the answer will be "shorter than what she measures it to be in her frame." Isn't that what you mean?

Another possible interpretation would be "shorter than itself." But that interpretation makes no sense to me, so I don't impute it to you. How can something be shorter than itself?

 

[aintnuthin]

One last way of approaching this:

GIVEN: Henry measures a distance to be 6LS in his frame. With respect to an object traversing that same distance at a speed of .6c with respect to him, TWO separate questions arise.

Question 1: Using the LT, what, according to Henry, will the other party measure that same distance to be in their frame?

Answer: 4.84 seconds. Why? Because he will assume that he is stationary, and will therefore expect the "moving" party to measure that same distance to be shorter. Their distance will be contracted, relative to him, or so he calculates, using the LT.

Question 2: Using the LT, and assuming that the other party calculates Harry's distance of 6 LS to be contracted, relative to themself, what will the other party (Harriet, here) calculate their own distance to be?

Answer: 7.5 light seconds. Compared to Harriet the "moving party" (Henry) will have the shorter distance (of 6 LS), as she calculates it.

They are two different questions, with two different answers, right?

 

[Doc Al]

Well, Doc Al seems to agree that I have the correct understanding:

Not exactly. You have the handwaving 'understanding' that lengths contract and clocks run slow, but you mix that up with more general statements about general distances and time intervals.

 

[Doc Al]

Doc, you say this is simple, but I can't see why. It seems abiguous to me, and it seems to me that this "simple" approach it is leading you to make inconsistent statements, which imply an infinite regression.

You say "use length contraction," but the unanswered question is "use it how?" Let's assume that you mean something like: "Whenever you see a length stated, in any context, then impute a lesser distance to all other frames." What results do we get?

No, that's not what I mean. If something is stationary in one frame, then you can use the simple length contraction formula to find its length (parallel to the direction of motion) as measured by some other frame. This doesn't apply to just any 'distance' measurement--for that you'd need the full LT.

Let's use a case where the relative speed is such (.866c, or whatever it is) that it results in a reduction of time and length of 50%. How does this work? Whose distance and time get reduced? Do the circumstances in which you came across the revelation of "length" matter at all?

Example 1: I say "a rocket is traveling at the rate of .866c (relative to Earth observers) to a star which is 8 light years away according to Earth observers"

So, you say: "stop right there. I've heard all I need to hear. That means the distance for the rocket is 4 light years."

Yes that's the distance between the stars according to earth. You can imagine a long pole stretching between the two stars that is stationary with respect to earth. (You can use the length contraction formula to find the distance according to the rocket.)

Example 2: I say "a rocket is traveling at the rate of .866c (relative to Earth observers) to a star which it deems to be 4 light years away."

The rocket frame measures the length of that 'pole' between the stars to be only 4 light years long. But in the rocket frame that 'pole' (the stars) is moving, thus Earth observers cannot use the simple length contraction formula on it.

So, you say: "stop right there. I've heard all I need to hear. That means the distance for Earth observers is 2 light years."

No, as I describe above, that is an incorrect application of the length contraction formula. You need to use the full LT. Then you'll get the correct result that Earth observers measure a distance of 8 light years.

Which is it? Do you ALWAYS just take whatever length you hear first, and then impute half that distance to the other party?

No, it's got to be the length of something that is stationary in one frame. That's the proper length of the object. All other frames see a shorter object.

 

[Dale]

It seems abiguous to me, and it seems to me that this "simple" approach it is leading you to make inconsistent statements

Only because you are too lazy to actually do a little math.

 

[Dale]

This conversation has become boring so I will work the problem.

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

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.

 

[ghwellsjr]

Post #24:

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?

Yes, it's wrong. To compute the interval between two events, use the distance and time measured within the same frame. You don't need to have 'proper' time or length.

Post #26:

This author (who is evidently a professor of physics), among others, seems to claim otherwise. After giving examples, a point he wants to stress is stated in this way:

"This demonstrates an important point: the observer who measures the proper length will not measure the proper time, and vice versa."

http://www.mta.ca/faculty/Courses/Physics/4701/EText/Proper.html

Do you disagree with him?

No, his statement is perfectly correct. But what does that have to do with your incorrect statement about mixing time and distance measurements from different frames to calculate an interval?

Do you know the definition of interval?

Doc Al, did you read the link that aintnuthin referenced? He only quoted the very end of it.

This same link came up in another thread and my comment was that it is total nonsense:

https://www.physicsforums.com/showpost.php?p=3407874&postcount=16

 

[Doc Al]

Doc Al, did you read the link that aintnuthin referenced? He only quoted the very end of it.

This same link came up in another thread and my comment was that it is total nonsense:

https://www.physicsforums.com/showpost.php?p=3407874&postcount=16

I was only agreeing with the final statement that was quoted. I agree that the link's discussion of 'proper time' is sloppy. (He's only talking of inertial frames, not the general case.) Thanks for the heads up.

 

[aintnuthin]

Doc Al, I have posed a number of questions, but, so far, at least, you have responded rather selectively. Is there a particular reason for that? For example:

1. Am I asking too many questions? More than you care (or have time) to answer?
2. Do you intend to answer them, later, but just don't have sufficient time right now?
3. Do some of my questions strike you as too incomprehensible and/or too ill-composed to respond to?
4. Do some of my questions strike you as being "too stupid" to even merit a response?

I appreciate your assistance, and I am willing to be patient and wait for your answers, if they will be forthcoming. Any answers you give me will, I hope, help me to better understand what you are saying, and may eliminate other questions.

 

[Dale]

Interesting how you are so willing to ask others to put in a lot of effort to answer a large string of questions, but so unwilling to put in even a minimal amount of effort of your own to do a single exercise.

 

[aintnuthin]

Interesting how you are so willing to ask others to put in a lot of effort to answer a large string of questions, but so unwilling to put in even a minimal amount of effort of your own to do a single exercise.

I ask questions here, realizing that no one is compelled to lift a finger trying to respond to any them, Dale, including you. Given the nature of your "answers," which to me are little more than unsupported assertions, without any attempt at providing a consistent, rational explanation for your claims, I'm not particularly interested in your "answers." I take the feeling to be mutual in that you're not interested in my questions.

The thrust of your comments have basically been in the vein of assailing my character (e.g. "lazy'), or my ability to understand. You will probably have more interest in, and get more entertainment out of, doing some math problems than addressing me. Math doesn't seem to "bore" you.

 

[Doc Al]

Doc Al, I have posed a number of questions, but, so far, at least, you have responded rather selectively. Is there a particular reason for that? For example:

1. Am I asking too many questions? More than you care (or have time) to answer?
2. Do you intend to answer them, later, but just don't have sufficient time right now?
3. Do some of my questions strike you as too incomprehensible and/or too ill-composed to respond to?
4. Do some of my questions strike you as being "too stupid" to even merit a response?

I appreciate your assistance, and I am willing to be patient and wait for your answers, if they will be forthcoming. Any answers you give me will, I hope, help me to better understand what you are saying, and many eliminate other questions.

How about this: Pick one set of questions, then wait for the answers. Don't stack up post after post. (It's all basically the same problem.)

I've answered quite a few of your questions. Are you still confused? If so, ask again.

My latest response is in post #47. I will await your comments on that before answering anything else.

 

[aintnuthin]

But in the rocket frame that 'pole' (the stars) is moving, thus Earth observers cannot use the simple length contraction formula on it...You need to use the full LT. Then you'll get the correct result that Earth observers measure a distance of 8 light years.

Is it sometimes impossible to use what you are calling the "full LT," or is it just that's it's unnecessary because convenient shortcuts are available?

No, it's got to be the length of something that is stationary in one frame. That's the proper length of the object. All other frames see a shorter object.

Your reference to the "length of an object" raises some questions in my mind, but I really don't want to get sidetracked on them right now. For now I have a question about this statement in particular: " All other frames see a shorter object."

Shorter than what? The very form of the word implies a comparison between two things, and therefore is impossible to interpret, standing alone.

 

[Dale]

Given the nature of your "answers," which to me are little more than unsupported assertions, without any attempt at providing a consistent, rational explanation

How is an explicit mathematical derivation from the Lorentz transform an "unsupported assertion" and not "a consistent, rational explanation" for this question? Your idea of "a consistent, rational explanation" needs some revision.

The thrust of your comments have basically been in the vein of assailing my character (e.g. "lazy'), or my ability to understand.

Yes, based on the evidence in this thread, I believe you are lazy. Regarding ability, I think you are perfectly able to understand, I am confident that the math is within your capability; you are just unwilling to make the effort required. Do you think that those of us who understand relativity did so without doing exercises such as the one I suggested for you?

I don't think you are wrong simply because you are lazy, I think you are wrong because I proved you are wrong. Similarly, you shouldn't think that I am wrong simply because I am arrogant and irritating. Look at the math, it is unambiguous.

 

[Doc Al]

Is it sometimes impossible to use what you are calling the "full LT,"

You can always use the full LT, if you know what you're doing and what the LT represents.

or is it just that's it's unnecessary because convenient shortcuts are available?

Exactly. For special cases, you can use 'shortcuts' such as time dilation and length contraction.

Your reference to the "length of an object" raises some questions in my mind, but I really don't want to get sidetracked on them right now. For now I have a question about this statement in particular: " All other frames see a shorter object."

Shorter than what? The very form of the word implies a comparison between two things, and therefore is impossible to interpret, standing alone.

It's simple. Say you are in a rocket that you measure to be 100 m long. You are traveling at .866c with respect to me. When I measure the length of your rocket, I get 50 m. Any frame moving with respect to you will measure a shorter length for your rocket than you do.

 

[aintnuthin]

You can always use the full LT, if you know what you're doing and what the LT represents... For special cases, you can use 'shortcuts' such as time dilation and length contraction.

OK, thanks. Would it be fair to assume, then, that using a shortcut (properly) would never give you an answer which was at odds with the one you would arrive at if you had taken the longer route and used the full LT (properly)?

It's simple. Say you are in a rocket that you measure to be 100 m long. You are traveling at .866c with respect to me. When I measure the length of your rocket, I get 50 m. Any frame moving with respect to you will measure a shorter length for your rocket than you do.

So, by implication, if the diameter of the Earth was, say, 8,000 miles, and I was in a rocket traveling .866c relative to you, then I would measure that diameter to be 4000 miles? Is that correct?

 

[Doc Al]

OK, thanks. Would it be fair to assume, then, that using a shortcut (properly) would never give you an answer which was at odds with the one you would arrive at if you had taken the longer route and used the full LT (properly)?

Right. It cannot be 'at odds' with it, since those shortcuts are derived from the full LT. But you must know when and where those 'shortcuts' can be applied.

So, by implication, if the diameter of the Earth was, say, 8,000 miles, and I was in a rocket traveling .866c relative to you, then I would measure that diameter to be 4000 miles? Is that correct?

Exactly.