How Does Relativity Affect the Measured Distance Between Cities at High Speeds?

[Abu]

Homework Statement

A man is flying with a speed of 0.7c and at a height 1 km. During his flight, he measures the distance between Toronto and Montreal to be 450 km. What is the real distance between them.

I don't think the answers are based on real life, they are just examples of places.

Homework Equations

t = t_o / √1-(v/c)²

The Attempt at a Solution

So if they are flying at 0.7c, that means they are flying at
0.7 x 3x10^8 = 2.1 x 10^8 m/s
Which means that they would cover 450 km, or 450000 meters, in 0.002142 seconds (I don't think the height of 1 km matters in this question, but correct me if I am wrong)

What I'm struggling with is the formula. I'm not sure where to put the numbers. From what I know, as you go faster, time increases slower. Since the man measured 0.002142 seconds, this is the dilated time, right? Meaning that this value of 0.002142 should be larger for an observer on earth. So for that reason, I want to put this value of 0.002142 as the t variable, instead of the t_o variable.

I think what is confusing me is which version of t is the dilated time. I've seen other versions of this time dilation formula without the division sign between t_o and the 1-(v/c)²

Thank you for your patience and time. If anything I said was not clear, please don't hesitate to ask me and I'll clarify my question.

Thank you!

 

[Chestermiller]

The distance between Montreal and Toronto is a length. Are you familiar with "length contraction?" if so, what is the equation for length contraction?

 

[Abu]

The distance between Montreal and Toronto is a length. Are you familiar with "length contraction?" if so, what is the equation for length contraction?

The equation for length contraction is:
L = L_o √1-(v/c)²

I didn't know that this formula could be applied to distances between places. I was under the impression that it only referred to the object that is moving. I thought it meant that Lo is the length of an object that is at rest, and that L was it's length once it was moving near the speed of light.

 

[Chestermiller]

The equation for length contraction is:
L = L_o √1-(v/c)²

I didn't know that this formula could be applied to distances between places. I was under the impression that it only referred to the object that is moving. I thought it meant that Lo is the length of an object that is at rest, and that L was it's length once it was moving near the speed of light.

If you are moving relative to the terrain below, it is the same as the terrain below moving relative to you (and you being stationary).

 

[Abu]

If you are moving relative to the terrain below, it is the same as the terrain below moving relative to you (and you being stationary).

I think I understand now... I saw a video that stated Lo is referred to as a proper length. The individual measures proper length when they are at rest to the object that they are measuring.

So in this instance, I am stationary relative to the object I am measuring (In this case, the object is the distance between Toronto and Montreal). The man on the other hand, is not stationary relative to the object he is measuring (still the distance between Toronto and Montreal).

That means that the answer should look like this:
L = L_o √1-(v/c)²
450000 meters = L_o√1-(0.7c/c)²
L_o = 630126 meters, or 630 kilometers
Is this right? If so, how come my reasoning regarding the time dilation formula was wrong... or is it possible and I was just doing it incorrectly?

Thank you for your time

 

[Abu]

No. This is not correct. Lo is 450 km, the distance in the rest frame of the object (which, in this case, is a line drawn on the surface of the Earth between Montreal and Toronto). As reckoned from the frame of reference of the man flying above the Earth at 0.7c, the distance is shorter.

Just to clarify... the question says that the man who is flying above Earth measures the distance between Toronto and Montreal to be 450 km. That is, from his reference frame the distance is 450 km, not shorter. Right?

Doesn't that mean that the real distance, which is the distance in the rest frame of the object, should be longer than 450 km?

Sorry if it seems like I am disregarding your responses, just processing the information...

 

[Janus]

No. This is not correct. Lo is 450 km, the distance in the rest frame of the object (which, in this case, is a line drawn on the surface of the Earth between Montreal and Toronto). As reckoned from the frame of reference of the man flying above the Earth at 0.7c, the distance is shorter.

According to the question, 450 km is the distance between Montreal and Toronto as measured by the man flying between them.

Just to clarify... the question says that the man who is flying above Earth measures the distance between Toronto and Montreal to be 450 km. That is, from his reference frame the distance is 450 km.

Doesn't that mean that the real distance, which is the distance in the rest frame of the object, should be longer than 450 km?

Sorry if it seems like I am disregarding your responses, just processing the information...

Yes, you are correct, the proper distance between Toronto and Montreal as measured in the ground frame would be longer than that measured by the man. (However, I would resist calling it the "real" distance, as the distance measured by the man is just as "real", just made in different frame of reference.)

 

[Chestermiller]

Sorry guys. I misread the question.

 

[Abu]

According to the question, 450 km is the distance between Montreal and Toronto as measured by the man flying between them.

Yes, you are correct, the proper distance between Toronto and Montreal as measured in the ground frame would be longer than that measured by the man. (However, I would resist calling it the "real" distance, as the distance measured by the man is just as "real", just made in different frame of reference.)

Ah okay, thank you two very much for your time. I greatly appreciate it.

After seeing your post I re-tried it using the time dilation formula, and so I did:
t = t_o / √1-(v/c)²
0.002142 = t_o / √1-(v/c)²
t_o = 0.00299
And then using the distance = speed x time formula
d = 2.1 x 10⁸ x 0.00299
d = 630 km (roughly)

So I got the right answer doing that method as well. My only remaining question is what does t and t_o truly mean?

Sorry guys. I misread the question.

It's no problem. I still really appreciate your time. Thank you.

 

[Janus]

Sorry guys. I misread the question.

Been there, done that

 

[David Lewis]

...what does t and to truly mean?

1. More convenient to use length contraction than time dilation in this case.
2. You don't have to convert speed to m/s. Just set c equal to 1.
3. I wouldn't use so many significant figures. The precision of the given data doesn't support it.

t is the time that elapses on the airplane when observed from Earth.
t₀ is the proper time -- how much time elapses in an observer's rest frame.
Note that t₀ > t

 

[Abu]

1. More convenient to use length contraction than time dilation in this case.
2. You don't have to convert speed to m/s. Just set c equal to 1.
3. I wouldn't use so many significant figures. The precision of the given data doesn't support it.

t is the time that elapses on the airplane when observed from Earth.
t₀ is the proper time -- how much time elapses in an observer's rest frame.
Note that t₀ > t

Thank you!

 

[Abu]

I just realized that what I did in post #9 was incorrect...

t is the time that elapses on the airplane when observed from Earth.
t₀ is the proper time -- how much time elapses in an observer's rest frame.
Note that t₀ > t

In the formula t = t_o / √1-(v/c)²
the √1-(v/c)² section is always less than one. That means that t > t_o right?
I really want to make the time dilation formula work for this problem. I read that t_o is the proper time, and that the proper time in a question is the time of the observer who is present at both destinations (this might be incorrect or not make sense, I'm not entirely sure). Which means that the proper time is not necessarily always the person who is at rest?

So that means that t_o in this problem is the man who is flying at 0.7c because he is present at both Toronto and Montreal, whereas the observer on Earth is simply watching the man fly. Is this correct?

EDIT**
Or is it just easier to identify who's time is dilated? Since the man is moving, his time is dilated, which means that his time value must be smaller than the Earth observer's time value, because time is passing slower for the flying man. What I mean is for example:
The man measures 5 seconds whereas the observer measures 10

 

[Chestermiller]

According to the clock of the guy on the plane, it takes 450/(0.7c)=to to travel from Montreal to Toronto. According to the clocks of the people on the ground, the interval between the guy's departure time at Montreal and his arrival time at Toronto is 630/(0.7c) = t.

 

[Abu]

According to the clock of the guy on the plane, it takes 450/(0.7c)=to to travel from Montreal to Toronto. According to the clocks of the people on the ground, the interval between the guy's departure time at Montreal and his arrival time at Toronto is 630/(0.7c) = t.

Right. So that means that for the formula:

T is always the time measured by the stationary observer
T_O is always the time measured by the moving object
And my reasoning is that T_O must be smaller than T because T_O's time is dilated. Time moves slower for the moving object, thus it's value for time must be smaller than that of the stationary object. Is this correct?

 

[Chestermiller]

Right. So that means that for the formula:
View attachment 219197
T is always the time measured by the stationary observer
T_O is always the time measured by the moving object
And my reasoning is that T_O must be smaller than T because T_O's time is dilated. Time moves slower for the moving object, thus it's value for time must be smaller than that of the stationary object. Is this correct?

I dislike interpretations like this. I prefer to think of it as a single observer with his clock at rest in his frame of reference, and multiple observers with their synchronized clocks at rest in their frame of reference.

 

[Janus]

Right. So that means that for the formula:
View attachment 219197
T is always the time measured by the stationary observer
T_O is always the time measured by the moving object
And my reasoning is that T_O must be smaller than T because T_O's time is dilated. Time moves slower for the moving object, thus it's value for time must be smaller than that of the stationary object. Is this correct?

The thing to remember is that which observer is the "Stationary" observer depends on which frame you are working from. In this scenario, If you are measuring from the Ground frame, then the ground observer is stationary and the man is moving. But in the Man's frame, he is stationary and the ground is moving.

Thus in the ground frame 0.003 sec passes during the trip, and the mans clock runs slow and ticks off 0.00214 sec during the trip.
However, for the Man, while he does measure 0.00214 sec passing for himself, he measures ground clocks as being time dilated and measures them as ticking off 0.00153 sec.
So how does this jive with the ground frame measuring 0.003 sec? There is a third effect besides length contraction and Time dilation to take into account. This is the Relativity of Simultaneity.
To explain how this works, imagine we have clocks on the ground in both Toronto and Montreal, and, in the ground frame, they are synchronized to each other (if one clock reads 0 the other clock simultaneously reads 0 ). However, in the Man's frame of reference, due to the Relativity of Simultaneity, if the Clock in Toronto reads 0 as he flies over it, the clock in Montreal does not read zero at that moment, but actually reads .00147 sec. The Montreal clock than advances .00153 sec to read 0.003 sec when he flies over Montreal. Thus both the Ground frame and Man agree that he passed over Toronto when Toronoto's clock read 0, and arrived at Montreal when Montreal's clock read 0.003 sec.

I don't know if you've covered the Relativity of Simultaneity yet or not, but it is the key to understanding a lot of SR scenarios. (Quite frankly, I think that any course of study dealing with SR should start with the Relativity of simultaneity, as a good grasp of this concept will avoid a lot of head-scratching later.)

 

[David Lewis]

Or is it just easier to identify whose time is dilated? Since the man is moving, his time is dilated...

Who is moving and who is at rest are conveniences for the mind, not features of the universe.