Does Time Dilation Affect Distance Measurements in Special Relativity?

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In summary: 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?The Lorentz transform is a mathematical model that describes the behavior of objects moving relative to each other. In this scenario, Henry is moving and Harriet is stationary. Therefore, according to the Lorentz transform, Harriet measures the distance between them as being shorter than what Henry measures. However, this does not mean that Harriet is the moving party - it is just a consequence of the way the Lorentz transform works.
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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.

I would appreciate any help you can offer.
 
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  • #2
One Brow said:
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.
Not correct.

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.
Correct.
 
  • #3
Doc Al said:
Not correct.


Correct.

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
aintnuthin said:
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?
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
Doc Al said:
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.


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
aintnuthin said:
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?
Yes.

If so, you have at least implicitly made Henry the stationary party, right?
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
Doc Al said:
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.

OK, thanks again for your answer. Your answer generates a few more questions, but let me start with this one.

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
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
aintnuthin said:
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.
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.)

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

Second question: If Henry "considered himself" to be stationary, what would Harriet's time and distance be from that perspective.
Henry does consider himself to be stationary. (And so does Harriet!)
My answer to that would be the one you gave, i.e., 8 seconds and 4.84 LS for Harriet.
That's the only answer there is as to what Harriet measures.

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?
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
aintnuthin said:
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 reasons:

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)
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
Doc Al said:
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.

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
Doc Al said:
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.

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
aintnuthin said:
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.
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.)

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?
See my comments above.
 
  • #14
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?
 
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  • #15
aintnuthin said:
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?
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
DaleSpam said:
Try actually doing a Lorentz transform.

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
aintnuthin said:
By hypothesis, Henry's time, in his frame is 10 seconds. By hypothesis, Henry's distance, in his frame, is 6 light seconds.
OK.
By hypothesis, Harriet is stationary and Henry is moving at .6c.
They are in relative motion; their relative velocity is 0.6c.

By virtue of the Lorentz Tranform, Harriet would have to see both is 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.
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
aintnuthin said:
Thanks for the suggestion, Dale, but I'm not a mathematician.
You don't have to be a mathematician, this is just middle-school level algebra. This excuse is not acceptable.
 
  • #19
aintnuthin said:
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.
So far, so good.
In this example, by hypothesis, Henry is the moving party.
Whether Henry is moving or not depends on who is the observer. Obviously to Harriet, Henry is moving.
So why doesn't Harriet see his time as slowed and his lengths contracted if she is stationary?
She does! She measures the length between the clocks to be contracted. She also measures Henry's clocks as running slow.
 
  • #20
Doc Al said:
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.

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?
 
  • #21
Doc Al said:
She does! She measures the length between the clocks to be contracted. She also measures Henry's clocks as running slow.

What does she measure his clock to read? 6.4 seconds?

If she reads only 8 seconds, and he reads 10 seconds, then her clock is slower, and his is faster, no?

Likewise, 4.84 light seconds is shorter, not longer, than 6 light seconds, right?
 
  • #22
aintnuthin said:
What does she measure his clock to read? 6.4 seconds?
She observes that during the time that she went from clock A to clock B, Henry's clocks show an elapsed time of 6.7 seconds. As far as she is concerned, those clocks run slow.

If she reads only 8 seconds, and he reads 10 seconds, then her clock is slower, and his is faster, no?
No. Her time is measured on a single clock, which allows Henry to apply time dilation to it. But the 10 seconds measured by Henry's clocks is a time interval measured on multiple clocks--you cannot apply the time dilation formula to it. (You can apply the LT, of course.) Remember that things are complicated by the relativity of simultaneity--which is part of the LT.

Likewise, 4.84 light seconds is shorter, not longer, than 6 light seconds, right?
Sure. Length contraction applies just fine.
 
  • #23
DaleSpam said:
You don't have to be a mathematician, this is just middle-school level algebra. This excuse is not acceptable.

Heh, Dale. I could easily turn that around. Even a first grader can express his thoughts isnwords. You can't?

I don't care to engage in a "flame war" with you. I merely asked if you could express your point in words. If you can't, then you can't. If you can, but don't care to, that's fine with me too.
 
  • #24
aintnuthin said:
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.
 
  • #25
Doc Al said:
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.

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 [Broken]

Do you disagree with him?
 
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  • #26
aintnuthin said:
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 [Broken]

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?
 
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  • #27
aintnuthin said:
Heh, Dale. I could easily turn that around. Even a first grader can express his thoughts isnwords.
I can express my thoughts in words: You are making wrong statements about what the Lorentz transform does, so you should actually work through the math to learn how they really work. One of the reasons that homework problems are such a key part of both math and physics is because of how much you learn by doing them.

Seriously, are you incapable of doing middle-school level algebra? If you have had algebra then the exercise will take you less than 20 minutes and you will learn a lot.
 
  • #28
Doc Al, you just said: "She observes that during the time that she went from clock A to clock B, Henry's clocks show an elapsed time of 6.7 seconds. As far as she is concerned, those clocks run slow."

Aren't we starting to go in circles here? If she measures his time to be 6.7 (sic) seconds (I calculate 6.4), then she would also measure his distance to be 3.872 light seconds, right? How is that the same as 6 light seconds, which is the only distance we were given at the outset of the problem?
 
  • #29
Doc Al said:
Do you know the definition of interval?

Maybe I don't. My understanding is that Minkowski came up with the ideas of proper time and proper lengths, and relies on those concepts heavily, in order to create something "invariant" in all frames. it is my further understanding that he could never possibly create such an invariant if he stuck to only one frame.

Do you agree with the claims made on this page, where the author points out that you can get either time dilation or time contraction with the same formula, depending on which way you make the measurement?

http://home.fnal.gov/~skent/cosmo/cosmo2.pdf (Starting near the bottom of page 2 with the paragraph beginning: "Another paradox of special relativity is the effect of "moving clocks run slow"...
 
  • #30
Doc Al said:
OK. They are in relative motion; their relative velocity is 0.6c.

They may be in relative motion, but we have, by hypothesis, postulation, and stipulation stated that Harriet is stationary. Agreed? Therefore we must stick with that, within the confines of this example.

Doc Al said:
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.

I agree, but disagree. I agree with you that 8 seconds (and 4.84 light seconds) would be what Henry calculated Harriet's time and distance to be IF he considered himself to be stationary.

However, I disagree to the extent that you are implying that, for the purposes of solving this particular problem, Harry can be viewed as stationary. By hypothesis, he is NOT stationary. He is moving, while Harriet is stationary. It makes a difference, doesn't it?
 
  • #31
aintnuthin said:
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".
 
  • #32
DaleSpam said:
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.
 
  • #33
aintnuthin said:
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.
 
  • #34
DaleSpam said:
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?
 
  • #35
aintnuthin said:
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.
 
<h2>1. What is time dilation in special relativity?</h2><p>Time dilation in special relativity is the phenomenon where time appears to pass slower for an observer in motion compared to an observer at rest. This is due to the fact that time and space are relative and can be affected by an object's velocity.</p><h2>2. How does time dilation affect distance measurements in special relativity?</h2><p>In special relativity, distance measurements are affected by time dilation because an object's velocity can cause time to pass slower for an observer, leading to a difference in the perceived distance between two objects. This means that the distance between two objects may appear shorter or longer depending on the relative velocity between them.</p><h2>3. Is time dilation a real phenomenon?</h2><p>Yes, time dilation is a real phenomenon that has been proven through various experiments and observations. It is a fundamental concept in special relativity and plays a crucial role in our understanding of the universe.</p><h2>4. How is time dilation calculated in special relativity?</h2><p>Time dilation is calculated using the Lorentz factor, which takes into account an object's velocity relative to the observer's frame of reference. The equation for time dilation is t' = t / √(1 - (v^2/c^2)), where t' is the perceived time, t is the actual time, v is the velocity, and c is the speed of light.</p><h2>5. Can time dilation affect everyday life?</h2><p>While time dilation is a fundamental concept in special relativity, its effects are only noticeable at very high velocities. In everyday life, the differences in time due to time dilation are extremely small and cannot be perceived without precise measurements. However, time dilation does play a crucial role in technologies such as GPS, which rely on precise time measurements for accurate navigation.</p>

1. What is time dilation in special relativity?

Time dilation in special relativity is the phenomenon where time appears to pass slower for an observer in motion compared to an observer at rest. This is due to the fact that time and space are relative and can be affected by an object's velocity.

2. How does time dilation affect distance measurements in special relativity?

In special relativity, distance measurements are affected by time dilation because an object's velocity can cause time to pass slower for an observer, leading to a difference in the perceived distance between two objects. This means that the distance between two objects may appear shorter or longer depending on the relative velocity between them.

3. Is time dilation a real phenomenon?

Yes, time dilation is a real phenomenon that has been proven through various experiments and observations. It is a fundamental concept in special relativity and plays a crucial role in our understanding of the universe.

4. How is time dilation calculated in special relativity?

Time dilation is calculated using the Lorentz factor, which takes into account an object's velocity relative to the observer's frame of reference. The equation for time dilation is t' = t / √(1 - (v^2/c^2)), where t' is the perceived time, t is the actual time, v is the velocity, and c is the speed of light.

5. Can time dilation affect everyday life?

While time dilation is a fundamental concept in special relativity, its effects are only noticeable at very high velocities. In everyday life, the differences in time due to time dilation are extremely small and cannot be perceived without precise measurements. However, time dilation does play a crucial role in technologies such as GPS, which rely on precise time measurements for accurate navigation.

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