Twins Paradox: Why One Twin is Older When Reunited

[Moris526]

Hi All. Layman question.

In the twin's paradox, why one of the brothers is older when they meet again?. if movement is relative. What determines which of them ends up older?

Hope i explained myself.

Thanks.

 

[ghwellsjr]

Hi All. Layman question.

In the twin's paradox, why one of the brothers is older when they meet again?. if movement is relative. What determines which of them ends up older?

Hope i explained myself.

Thanks.

You explained your question very well. The answer is the Principle of Relativity and the Principle that the speed of light is independent of the speed of the source of the light, both of which have been experimentally verified.

This means that as the brothers are separating at a constant speed, they will each see the other one's clock as running slower than their own by exactly the same ratio. We'll call this ratio R. Then when one of the brothers turns around and approaches his brother at the same speed going the opposite direction, he will see his brothers clock running faster than his own by exactly the inverse of the first ratio, which is 1/R. Since he spends the same amount of time traveling away as traveling back, we can average these two ratios to get the final ratio of his brother's accumulated age to his own accumulated age. The average will be (R+1/R)/2. If this average is greater than one, then the twin who turned around sees his brother is older, correct?

Why don't you pick any value of R less than 1 and see what you get?

Does that answer your question?

 

[ghwellsjr]

I should also explain why the ratios for going and coming at the same speed are the inverse of each other.

Let's say that you are looking at two observers with clocks. One of them is a fixed distance away from you so you will see his clock ticking at the same rate as your own. They may have different times on them but we don't care about that, we only care about tick rates. The second observer is traveling at a constant speed from the first one directly toward you. Therefore, you will see his clock ticking faster than yours. You also know that you are seeing time progress on the traveler's clock compared to the first observer's clock just as the traveler is seeing them but delayed in time, provided that the light from both clocks travels towards you at the same speed. Therefore the ratio of the tick rates that you see of the distant fixed clock to the traveler's clock is some ratio that we called R. But the ratio of the tick rates that you see of the traveler's clock compared to your own (which is the same as the distant fixed clock) is the inverse of R.

Knowing that the speed that the traveler is going away from the distant fixed clock is the same as the speed that the traveler is approaching you wraps up the proof that we are looking for.

Does that make sense?

 

[Popper]

Hi All. Layman question.

In the twin's paradox, why one of the brothers is older when they meet again?. if movement is relative. What determines which of them ends up older?

Hope i explained myself.

Thanks.

The twin paradox refers to one twin staying at home while the other twin travels to a planet, turns around an comes back to where it started. Suppose that one twin is on eath and at t = 0 the other twin is moving past the stay at home twin at a significant fraction of the speed of light. This means that their motion is not symmetric since each twin has different experiences. The twin that goes away and comes back must accelerate during his journey and its this acceleation which breaks the symmetry which existed between the two twins. The twin who is accelerating can think of himself as being in a gravitational field and clocks higher up in a gravitational field run at different rates than ones lower in the gravitational field. The accelerating twin can think of himself as being at the place where the potential energy of any object is zero. The clock which is with his twin back on Earth (perhaps his wrist watch or his biological clock) will run at a different rate than his is. When all is said and done the calculations will show that the stay at home twin aged less.

Essentially it all boils down to this: The worldline of the stay at home twin is a straigt line while the twin on the journey isn't straight. You can idealize the accelerating twins worldline as two straight line segments, one of the ends of one segement is at the event "traveling twin is at same location a stay at home twin" while the other end of the segent is at the event "traveling twin turns around and heads for home." The second segment of his worldline starts at this last event and eneds at the event "traveling twin comes back home to meet his twin." Draw these worldlines on a spacetime diagrame and you'll see that one is straight and the other is a bent line where the bend is sharp bend, the sharper the bend the larger the accelertion of the traveling twin.

 

[Moris526]

Thank you both for your replies.

But i still don't get something.

If the twin on the ship accelerates away, turns back and accelerates again, doesn't that mean that from his/her perspective his/her brother at Earth does the same thing?. I don't understand what differentiates one of the frames of references. We could as well say that one brother stays on planet Ship and the other flies away on ship earth.

Or that's just a mathematical consequences and won't make common sense?

Thanks again.

Ive always liked physics, never could study. So now I am trying to learn something on my own.

 

[PeterDonis]

I don't understand what differentiates one of the frames of references.

One twin, the one we call the "traveling" twin, has to fire rockets to turn around. He feels acceleration when he does that. The other twin, the one we call the "stay-at-home" twin, doesn't do anything; he just floats freely the whole time, and feels no acceleration. That's the difference between the two.

 

[ghwellsjr]

Thank you both for your replies.

But i still don't get something.

If the twin on the ship accelerates away, turns back and accelerates again, doesn't that mean that from his/her perspective his/her brother at Earth does the same thing?. I don't understand what differentiates one of the frames of references. We could as well say that one brother stays on planet Ship and the other flies away on ship earth.

Or that's just a mathematical consequences and won't make common sense?

Thanks again.

. I've always liked physics, never could study. So now I am trying to learn something on my own.

Did you do the calculation I asked you to do to figure out which one aged more? I don't see that you have gotten an answer to your question. What is the correct answer?

And please keep in mind that my answer to your question does not require you to identify any frame of reference but even if you want to do that, it won't matter which frame of reference you pick, they all will give the same answer so your statement that you don't understand what differentiates one of the frames of reference doesn't have any bearing on the answer to your question.

 

[Moris526]

One twin, the one we call the "traveling" twin, has to fire rockets to turn around. He feels acceleration when he does that. The other twin, the one we call the "stay-at-home" twin, doesn't do anything; he just floats freely the whole time, and feels no acceleration. That's the difference between the two.

Thanks. I get it now.

 

[ghwellsjr]

Thanks. I get it now.

OK, so you understand that there is a difference between the twins, but how does that lead to an answer to your question? Have you determined which twin ends up older? What is your answer?

 

[Samshorn]

I should also explain why the ratios for going and coming at the same speed are the inverse of each other. Let's say that you are looking at two observers with clocks. One of them is a fixed distance away from you so you will see his clock ticking at the same rate as your own. They may have different times on them but we don't care about that, we only care about tick rates. The second observer is traveling at a constant speed from the first one directly toward you. Therefore, you will see his clock ticking faster than yours. You also know that you are seeing time progress on the traveler's clock compared to the first observer's clock just as the traveler is seeing them but delayed in time, provided that the light from both clocks travels towards you at the same speed. Therefore the ratio of the tick rates that you see of the distant fixed clock to the traveler's clock is some ratio that we called R. But the ratio of the tick rates that you see of the traveler's clock compared to your own (which is the same as the distant fixed clock) is the inverse of R.

That explanation doesn't seem very clear to me. I think a clear explanation would have to convey why the frequency ratios are inverses according to special relativity, even though they would not be inverses according to classical Doppler. This difference is the key factor that distinguishes special relativity from pre-relativistic physics, and of course it's also key to understanding the twins paradox, but I don't see an explanation of the difference here. Also it seems confusing and unnecessary to introduce a third observer/clock that is stationary relative to one of the other two observer/clocks... but perhaps it just seems that way because I don't follow your explanation.

By the way, in order to explain why "the ratios for going and coming at the same speed" are reciprocals, I think you will need to specify how you define "speed". You can't base it on the Doppler ratios, because that would be circular. You need some independent means of defining the "speed" of the twin. Normally this is done by defining a suitable system of space and time coordinates. If you think you can dispense with this, and simply (for example) define equal outgoing and incoming speeds as those for which the Doppler ratios are reciprocal, then the proposition you're trying to prove is actually tautological... but then you face the task of showing that this definition of "speed" agrees with the normal definition of speed - which of course is based on a suitable system of space and time coordinates, the very thing you are trying to do without.

 

[Moris526]

OK, so you understand that there is a difference between the twins, but how does that lead to an answer to your question? Have you determined which twin ends up older? What is your answer?

I didn't understand why should any be older than the other if movement is relative and there were no difference between the frames of reference. Now, the acceleration of the ship makes a difference between them. Thou i have to admit i can't fully visualize why the Earth man ends up older.

I have the notion that time slows down as you accelerate but i can't make that clear in my head. I don't kown if it can be done or is just a logical conclusion of the math.

You should also know that i have just basic knowledge of mathematics.

By the way, thanks for your time.

 

[ghwellsjr]

That explanation doesn't seem very clear to me. I think a clear explanation would have to convey why the frequency ratios are inverses according to special relativity, even though they would not be inverses according to classical Doppler.

You must have understood my explanation enough to determine that there is no difference between Relativistic Doppler and Classical Doppler for what I described and I agree.

This difference is the key factor that distinguishes special relativity from pre-relativistic physics, and of course it's also key to understanding the twins paradox, but I don't see an explanation of the difference here. Also it seems confusing and unnecessary to introduce a third observer/clock that is stationary relative to one of the other two observer/clocks... but perhaps it just seems that way because I don't follow your explanation.

I should have carried the explanation two steps further and applied the Principle of Relativity to point out that whatever Doppler ratio you see of the traveling observer's clock coming toward you compared to your clock, he sees the exact same Doppler ratio of your clock to his. And whatever Doppler ratio the traveling observer sees of the remote stationary clock to his, is the same that the remote observer sees of the traveling observer's clock to his own. In other words, there are only two ratios to consider among the three observers (not counting the unity ones between you and the remote observer). In Classical Doppler, there can be four.

By the way, this little exercise is not part of the Twin Paradox scenario--it is just meant to be an independent proof that the Doppler ratios are inverses for approaching and retreating at the same speed.

By the way, in order to explain why "the ratios for going and coming at the same speed" are reciprocals, I think you will need to specify how you define "speed". You can't base it on the Doppler ratios, because that would be circular. You need some independent means of defining the "speed" of the twin. Normally this is done by defining a suitable system of space and time coordinates. If you think you can dispense with this, and simply (for example) define equal outgoing and incoming speeds as those for which the Doppler ratios are reciprocal, then the proposition you're trying to prove is actually tautological... but then you face the task of showing that this definition of "speed" agrees with the normal definition of speed - which of course is based on a suitable system of space and time coordinates, the very thing you are trying to do without.

I'm not defining the speed of the traveling twin nor am I determining what the relationship is between speed and Doppler ratio--only that they are inverses for approaching and retreating at the same speed. I didn't even think it was necessary to state how the traveling twin determines that he is traveling in both directions at the same speed--that could be a mere assumption. But in any case, the traveling twin could legitimately use the inverse Doppler ratio to determine that his speed was the same on the return trip as on the outgoing (without identifying what that speed is) and it is important that the speeds be equal in order for the travel times to be equal so that the calculation can be legitimate.

Keep in mind, I'm only trying to answer the OP's question: which twin ages more?

 

[ghwellsjr]

I didn't understand why should any be older than the other if movement is relative and there were no difference between the frames of reference. Now, the acceleration of the ship makes a difference between them.

Do you understand that looking at any activity while traveling away from it will make it appear to be in slow motion and while approaching, it will make it appear in fast motion? And do you know that this is called Doppler?

Thou i have to admit i can't fully visualize why the Earth man ends up older.

Are you sure the Earth man ends up older? Popper said the opposite:

When all is said and done the calculations will show that the stay at home twin aged less.

I have the notion that time slows down as you accelerate but i can't make that clear in my head.

You don't have to know about time slowing down or anything about acceleration to know how to get the answer to your question.

I don't kown if it can be done or is just a logical conclusion of the math.

You should also know that i have just basic knowledge of mathematics.

By the way, thanks for your time.

Was the math of post #2 within your grasp?

 

[Samshorn]

I should have carried the explanation two steps further and applied the Principle of Relativity to point out that whatever Doppler ratio you see of the traveling observer's clock coming toward you compared to your clock, he sees the exact same Doppler ratio of your clock to his. And whatever Doppler ratio the traveling observer sees of the remote stationary clock to his, is the same that the remote observer sees of the traveling observer's clock to his own.

Yes, by invoking the principle of relativity along with the independence of light speed from the speed of the source, I agree that you can infer that the Doppler shift observed by the traveling twin on his outbound journey is the reciprocal of the shift on his return journey (with symmetrical speeds). And from this it follows that the total elapsed time for the traveling twin is less than for the stay-at-home twin. If I was to paraphrase your argument, I'd say you are considering three clocks a,b,c, with a and c mutually at rest, with clock b directly in between a and c, and moving away from a and toward c. Letting Da/Db denote the ratio of clock ticks from a reaching b to clock ticks of b, and so on, we have (Da/Db)(Db/Dc) = Da/Dc = 1, and by relativity we have Dc/Db = Db/Dc. Therefore, Da/Db is the reciprical of Dc/Db, and these represent the Doppler shift ratios for approaching and receding at the speed of b. Then, letting R denote the receding Doppler ratio, and 1/R the approaching Doppler ratio, you point out that for a symmetrical trip the average Doppler ratio for the whole trip is (R + 1/R)/2, which is strictly greater than 1 as R differs from 1. Hence the traveling twin ages less.

You must have understood my explanation enough to determine that there is no difference between Relativistic Doppler and Classical Doppler for what I described and I agree.

I don't follow that. Classical Doppler (for a theory in which the speed of the wave is independent of the speed of the source, like sound waves) would not lead to the same result, because classical Doppler gives different frequency ratio for the cases when the source or the receiver is moving.

Also I think this line of reasoning doesn't enable us to quantify the magnitude of the relativistic time dilation for any specific trip velocity, since it doesn't relate the speed to the Doppler ratios.

Keep in mind, I'm only trying to answer the OP's question: which twin ages more?

Yes, although I think it's debatable how much this explanation really clarifies why the twin who turns around ages less. It shows that if we impose the two principles, the turn-around twin must age less, but for a beginning student those two principles seem to be irreconcilable, and everyone knows we can derive infinitely many false conclusions from self-contradictory premises. So, for this reasoning to be persuasive, we need to explain why those two principles are not mutually contradictory (as they would be classically). And of course the light-speed principle applies only to inertial coordinates, so it comes back to the question of what makes one path inertial and the other not (if we imagine a purely relational world, for example, although the OP doesn't seem to be concerned about that - yet). This, combined with the fact that this approach doesn't enable us to quantify the effect, may explain why this approach is not often followed.

 

[ash64449]

Are you sure the Earth man ends up older? Popper said the opposite:
When all is said and done the calculations will show that the stay at home twin aged less.

Well,according to Wikipedia It is opposite of what popper said!

http://en.wikipedia.org/wiki/Twin_paradox

"""In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more."""

 

[Dale]

Well,according to Wikipedia It is opposite of what popper said!

http://en.wikipedia.org/wiki/Twin_paradox

"""In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more."""

The inertial twin is always the one with the maximum proper time (i.e. stay at home twin ages more). I suspect Popper knew that and just wrote it wrong.

 

[ash64449]

The inertial twin is always the one with the maximum proper time (i.e. stay at home twin ages more). I suspect Popper knew that and just wrote it wrong.

How is maximum proper time indicate that time dilation is less?(just curious)

 

[Dale]

How is maximum proper time indicate that time dilation is less?(just curious)

For me, the best way to keep it straight is to think of it geometrically. Think about Euclidean geometry. Given two points you can connect them with an infinite number of paths, each of which will have a certain length. The shortest length between the two points is a straight line, and all other lines will be bent and have a longer length.

In spacetime a similar idea holds, except that now there are different kinds of paths, timelike paths whose lengths are measured by clocks and spacelike paths whose lengths are still measured by rods. The other difference is that for timelike paths the "straight line" statement gets flipped so the LONGEST time between two events is a straight line, and all other lines will be bent and have a SHORTER time.

 

[ash64449]

For me, the best way to keep it straight is to think of it geometrically. Think about Euclidean geometry. Given two points you can connect them with an infinite number of paths, each of which will have a certain length. The shortest length between the two points is a straight line, and all other lines will be bent and have a longer length.

In spacetime a similar idea holds, except that now there are different kinds of paths, timelike paths whose lengths are measured by clocks and spacelike paths whose lengths are still measured by rods. The other difference is that for timelike paths the "straight line" statement gets flipped so the LONGEST time between two events is a straight line, and all other lines will be bent and have a SHORTER time.

hey,, that means...

An hour is closer to me than a minute?!

 

[Dale]

Huh?

 

[Nugatory]

hey,, that means...

An hour is closer to me than a minute?!

No. Consider three events ("event" is just another word for "point in space-time"):
1) You clap you hands right now. When you do, your wristwatch reads exactly noon. Call this event N.
2) You clap you hands again. This time your wristwatch reads 12:01. Call this event M.
3) You clap your hands a third time. This time your wristwatch reads 13:00. Call this event H.

Assume that you're just sitting still, no acceleration or other change of speed, for the entire hour. It might be best to assume that you're floating in empty space, so we don't have to worry about gravity, the rotation of the Earth, and the like.

Now N, M, and H all lie on the same straight line; M lies between N and H; and as you travel through space-time you are moving along that line, starting at point N, passing through M, and then reaching H.
Clearly M is closer to you than H because you had to pass through M to get to H on a straight-line path. Your path from N to M is one minute long, your path from M to H is 59 minutes long, and your path from N to H through M is one hour long.

However, there are paths through space-time from N to H that don't pass through M. Your path does, because all three events are you clapping your hands, so they're always going to happen where you are.
But consider a spaceship that starts out sitting right next to you, takes off when you clap your hands for the first time, flies for a while, turns around, and returns just in time to land and be sitting next to you when you clap your hands for the third time.
The spaceship's path through spacetime passes through N and H but not through M. It's also not a straight line.

DaleSpam is saying that the spaceship's path between N and H will be shorter than your path between N and H - not too surprising because they are different paths. That has nothing to do with the event M which is on your path but not the spaceship's. Your path from N to M is one minute long, your path from M to H is 59 minutes long, your path from N to H through M is one hour long, and for you H is further away than M.

 

[ash64449]

nugatory,yes. You are right.. Makes sense..
I got exicted and searched in google and found a video just what i said. Unfortunately,i can't paste the url of the video.
Let me tell the description of the video and tell me what you feel from that video
video is from minutephysics.
Name of that video:Distance and Special Relativity:
How far away is tomorrow?

 

[ash64449]

Huh?

i will tell you.
You said the shortest 'space like' intervel(straight line) is considered longest time between two events.(proper time-time like). Hour is longest time than a minute.so in space like intervel,hour is shortest distance than a minute so hour is closer to me than a minute! Dosen't that make sense.
I figured this out and got exicted and googled this thing and found a video.unfortunately,i am not able to paste the video's url.i will tell the description so that you can see it by yourself.
Video is from minutephysics
name of video:Distance and Special Relativity:
How far away is tomorrow?

 

[ash64449]

nugatory and dalespam:
can you guys explain me in simpler manner why shortest 'space like' intervel is considered as longest 'time like' intervel?

 

[ash64449]

and i think why we can get to hour faster than the minute is because whenever we move,we move through multiple dimensions!
That is we move through spatial dimension and time like dimension at the same time..

 

[BruceW]

nugatory and dalespam:
can you guys explain me in simpler manner why shortest 'space like' intervel is considered as longest 'time like' intervel?

This sentence is not correct in several ways. You need to know what the terms mean.
'spacetime' is just a word to show that we are talking about a concept in Einstein's relativity
'spacetime interval' Well, the interval in classical mechanics is just the distance between two points. This 'spacetime interval' is a similar concept in relativity. So it sort of represents a kind of 'distance', but it is not the same as distance as in classical mechanics. To be specific, this is the equation for the spacetime interval between two points (in 1d, with zero gravity):
$$\int \sqrt{c^2dt^2 - dx^2} = spacetime \ interval$$
so if you have dx=0 then the spacetime interval is going to be as great as possible. This corresponds to the person that does not move. So the person that does not move has a greater spacetime interval than the person that flies away on a ship and comes back again.
'time like' this is any path that is going at less than light-speed (i.e. all physical objects move on a time like path).
'space like' this is any path that is not 'time like', or moving at the speed of light. 'space like' paths are not important for the twin paradox problem.
'proper time' this means the same as the 'spacetime interval'
Also, the 'proper time' for a given path is the time which a person traveling on the path would measure on their watch. (or equivalently, how much they age). So from the above reasoning, the person that ages the most is the person who stays at home on the planet earth.

Finally, keep in mind that sometimes the spacetime interval is defined with the minus sign the other way around, and the units are sometimes defined differently.

 

[Nugatory]

To be specific, this is the equation for the spacetime interval between two points (in 1d, with zero gravity):
$$\int \sqrt{c^2dt^2 - dx^2} = spacetime \ interval$$

All I would add is that if someone prefers a calculus-free formula... as long as you're also working with velocities that are constant or change instantaneously (for example in the most common statement of the twin paradox we say that the traveling twin goes from 0 to v instantaneously at the start of the journey then travels at a constant v until the turnaround point) then this integral turns into:$$s^2 = (t_2-t_1)^2 - (x_2-x_1)^2$$
where s is the spacetime interval and the t's and x's are the coordinates of the two events; it doesn't matter whose coordinates we use, the interval comes out the same.

This works for any spacetime diagram that only has straight lines, no curves, in it.

 

[PeterDonis]

Was the math of post #2 within your grasp?

It's not just the math of post #2; the OP needs to understand why the math of post #2, as you presented it, only applies to the traveling twin, not the stay-at-home twin. There is also a corresponding equation that applies to the stay-at-home twin, with the same ratio R in it, but it differs from the one you gave in post #2 in a crucial respect. What that crucial respect is is not a question of math, it's a question of physics. Post #2 gives a hint at the answer to the question of physics, but you didn't give it explicitly.

 

[Dale]

You said the shortest 'space like' intervel(straight line) is considered longest time between two events.(proper time-time like).

No. I said the longest timelike interval between two events is a straight line. I never mixed up timelike and spacelike intervals.

The shortest spacelike interval between two events is a straight line.
The longest timelike interval between two events is a straight line.

Those are two completely separate statements about different pairs of events and different types of spacetime interval. I don't know why you are mixing them up like that.

Hour is longest time than a minute.so in space like intervel,hour is shortest distance than a minute so hour is closer to me than a minute! Dosen't that make sense.

No. You can certainly use geometrized units where time is measured in units of distance, but that doesn't make an hour closer than a minute. In such units an hour would be 1E9 km and a minute would be 1.8E7 km. So a minute is a shorter distance than an hour.

How far away is tomorrow?

2.6E10 km

 

[Dale]

nugatory and dalespam:
can you guys explain me in simpler manner why shortest 'space like' intervel is considered as longest 'time like' intervel?

It isn't. You are mixing up spacelike and timelike intervals. You never compare timelike and spacelike intervals.

 

[ghwellsjr]

This sentence is not correct in several ways. You need to know what the terms mean.
'spacetime' is just a word to show that we are talking about a concept in Einstein's relativity
'spacetime interval' Well, the interval in classical mechanics is just the distance between two points. This 'spacetime interval' is a similar concept in relativity. So it sort of represents a kind of 'distance', but it is not the same as distance as in classical mechanics. To be specific, this is the equation for the spacetime interval between two points (in 1d, with zero gravity):
$$\int \sqrt{c^2dt^2 - dx^2} = spacetime \ interval$$
so if you have dx=0 then the spacetime interval is going to be as great as possible. This corresponds to the person that does not move. So the person that does not move has a greater spacetime interval than the person that flies away on a ship and comes back again.
'time like' this is any path that is going at less than light-speed (i.e. all physical objects move on a time like path).
'space like' this is any path that is not 'time like', or moving at the speed of light. 'space like' paths are not important for the twin paradox problem.
'proper time' this means the same as the 'spacetime interval'
Also, the 'proper time' for a given path is the time which a person traveling on the path would measure on their watch. (or equivalently, how much they age). So from the above reasoning, the person that ages the most is the person who stays at home on the planet earth.

Finally, keep in mind that sometimes the spacetime interval is defined with the minus sign the other way around, and the units are sometimes defined differently.

This is all wrong, unless somebody changed the definition of Spacetime Interval and didn't tell me. Where did you get that definition?

As far as I know, the Spacetime Interval is not an integral and therefore does not depend on the path between two points. A correct formula from wikipedia is:

(spacetime interval)
​

If s² evaluates to a negative number, then the spacetime interval is timelike and can be measured by a clock at rest in an inertial frame. This is also the Proper Time on the clock but that is immaterial.

If s² evaluates to a positive number, then the spacetime interval is spacelike and can be measured by a ruler at rest in an inertial frame. This is also the Proper Distance on the ruler but that is immaterial.

If s² evaluates to zero, then the spacetime interval cannot be measured but is defined to be the propagation of light in any inertial frame.

 

[PeterDonis]

As far as I know, the Spacetime Interval is not an integral and therefore does not depend on the path between two points.

In flat spacetime, yes, you can define the "spacetime interval" as you have defined it; but the definition doesn't generalize to curved spacetime. (It also doesn't generalize to non-inertial frames in flat spacetime; your definition requires that the coordinates are defined relative to an inertial frame.)

The integral definition is more general in that it works in curved spacetime and in non-inertial frames in flat spacetime; but, as you note, it makes the "interval" path-dependent, because you have to choose a curve along which to integrate. If you choose a geodesic curve, the integral definition is equivalent to yours, but, as I said, it works in curved spacetime and in non-inertial frames in flat spacetime. If you choose some non-geodesic curve, the integral will give you a number, which may have a physical meaning (for example, it gives the elapsed proper time along a non-geodesic timelike curve), but will be different than the integral evaluated along a geodesic.

AFAIK the term "spacetime interval" is only used for the length along a geodesic in flat spacetime, which matches your definition; but that just means the term "spacetime interval" is of very limited application. The integral form, since it's more general, is much more useful, IMO.

 

[ash64449]

It isn't. You are mixing up spacelike and timelike intervals. You never compare timelike and spacelike intervals.

yes. Sorry,i think i did mix them up...

 

[ash64449]

It isn't. You are mixing up spacelike and timelike intervals. You never compare timelike and spacelike intervals.

But DaleSpam,you know what i meant...

Why is Longest Time like interval a straight line?

 

[Nugatory]

Why is Longest Time like interval a straight line?

This follows from the structure of spacetime, where the interval between two events is given by ${\Delta}s^2 = {\Delta}t^2 - {\Delta}x^2$. Try plugging in some values for Δx and Δt (remember that we're talking about timelike intervals, so Δt is always greater than Δx, and I chose the sign convention to make it easy to work with timelike intervals) and you'll see that the straight line path between two points A and B is always longer than the unstraight path that passes through some third point C.

This is different from ordinary space where the interval between two points is given by ${\Delta}s^2 = {\Delta}x^2 + {\Delta}y^2$ (just the familiar Pythagorean theorem for calculating the distance between points in the x-y plane) where the straight line is always the shortest distance.