# Why does time slow when you travel at high speeds?

I can't really bring myself to understand why time slows when you travel at high speeds. I tried to think of a graph in witch the y is time and the x is distance. This line could be at any positive slope. So the only relation is the "speed" at witch the line moves on the graph.

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Time does NOT slow when you travel at high speeds. Right now, as you are reading this, you are traveling at .9999c Does your time feel slowed down?

The APPEARANCE of time slowing down is an artifact of remote observation from a different inertial frame of reference.

MathJakob
I can't really bring myself to understand why time slows when you travel at high speeds. I tried to think of a graph in witch the y is time and the x is distance. This line could be at any positive slope. So the only relation is the "speed" at witch the line moves on the graph.

I'm sure the professionals will jump on me for giving you a basic although not entirely accurate explanation but this is accurate enough and you will understand.

Space and time are interlinked. I have two identical stationary objects, for both of those objects, time is exactly the same. For every second that passes for one object, one second will pass for the other object. Now if we accelerate one of the objects, even by a small amount, they will have different frames of reference.

This means that for the stationary object, when one second passes, 0.999999999 of a second will pass for the object that is moving (example figures)

So time for the moving object is slower relative to the stationary object. Now if you accelerated the other object to the same speed, they would both have the same frame of reference again. Their times are synced if you will. They are the same.

How do we know this is true aside from the equations? Well the people inside international space station have to adjust their atomic clocks ever so slightly, and I mean the tinest little bit so they stay synced with the clocks on Earth. This is because the ISS is moving at 7.71km/s so the ISS's frame of reference is different to Earths. Also it's gravity is different but I won't go into that.

So technically when you walk down the street and pass someone who is standing still, or you're in a car and you overtake another car, time is slower for you, although the amount is so incredibly small that it's impossible to notice or even measure (probably)

Take two identical cars, the one that is moving weighs more than the stationary one, because your mass increases the faster you go, again the difference is immeasurable. So what would happen if you went really really fast compared to someone who was stationary?

Let's say you're sitting in a train that is circling the Earth at near the speed of light, you're windows are blacked out and the only thing you can see is what is inside your train (the moving train is your frame of reference). Everything would appear normal, you would get up, walk around, drink, eat, talk, breath and everything would appear normal, because everything inside the train is inside the same frame of reference, but the stationary people outside on the platform looking in as you whiz past would see something totally different, they would see you barely moving, you would be in super mega slow motion because relative to the people outside the train, you're moving in slow motion.

Assume you are sitting at the table on the train and the people outside were looking in as you whiz past, they'd have to watch the train for 1 week their time, just to see you lift a glass of water to your mouth (example figures) because of the time difference, but you don't notice this... as far as you're concerned it took you 2 seconds to lift the glass and take a sip of water.

Finally, if you looked out the window at the people who are moving at normal speeds, walking driving ect, it would look like they're whizing all over the place at super speeds, again this is just the time difference.

I hope I didn't confuse you too much and I'll probs get jumped on for any inaccuracies I made but I hope you get the general idea.

yuiop
Assume you are sitting at the table on the train and the people outside were looking in as you whiz past, they'd have to watch the train for 1 week their time, just to see you lift a glass of water to your mouth (example figures) because of the time difference, but you don't notice this... as far as you're concerned it took you 2 seconds to lift the glass and take a sip of water.

Finally, if you looked out the window at the people who are moving at normal speeds, walking driving ect, it would look like they're whizing all over the place at super speeds, again this is just the time difference.
I am not an expert and it is not my intention to jump all over you, but that last statement is the opposite of what would be observed. The people on the platform will be whizzing past from the point of view of the observers on the train, simply because of the relative motion, just as the training is whizzing past from the point of view of the observers on the platform. However, when a person on the train observes somebody on the platform lifting a glass of water they will also have to wait a week to see them complete that action. The time dilation effect is completely symmetrical.

xcourrier
As far as I know, the best answer to this is because the speed of light is constant to all observers.

If you are sitting down in a cafe and a car drives by at 30 mph, the driver and everything in that car is moving at 30 mph to you. If someone throws a ball from that car in the direction they are going at 40 mph, that ball will be traveling at 70 mph to you (assuming no air resistance). The speed of the ball is not constant to all observers, to you its traveling at 70 mph and to the people in the car it is traveling at 40 (to the people in the car, you are traveling at -30mph and they are remaining stationary).

Light is the only thing that is constant for all observers. If the car turns on its headlights, the driver measures those photons traveling away from him at 3x10^8 meters per second. You measure those photons at 3x10^8 meters per second as well; not at 3x10^8 m/s + 30 mph, like the situation with the ball. Space and time have to change to keep it constant for both of you.

<-good overview

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MathJakob
I am not an expert and it is not my intention to jump all over you, but that last statement is the opposite of what would be observed. The people on the platform will be whizzing past from the point of view of the observers on the train, simply because of the relative motion, just as the training is whizzing past from the point of view of the observers on the platform. However, when a person on the train observes somebody on the platform lifting a glass of water they will also have to wait a week to see them complete that action. The time dilation effect is completely symmetrical.

I was not aware of that, can someone else confirm? Thanks.

xcourrier
I was not aware of that, can someone else confirm? Thanks.

That is correct. Regardless of how you are moving, you can always assume yourself stationary and the rest of the world moving.

When someone sees you moving really really fast in a train, it looks to them like you're moving slowly. However, to you the rest of the world is moving really really fast and their time is going slow while you experience time normally.

Gold Member
I can't really bring myself to understand why time slows when you travel at high speeds. I tried to think of a graph in witch the y is time and the x is distance. This line could be at any positive slope.
No, you can't travel on a line at any positive slope, the slope must be greater than 1 (and it can also be more negative than -1). It just can't be 1 because that is the speed of light and slopes smaller than that would be faster than light.

So the only relation is the "speed" at witch the line moves on the graph.
A vertical line represents an object that is stationary in the frame. That's because its x coordinate doesn't change as its t coordinate does change. A slightly off vertical line represents an object moving at a slow speed (either positive or negative). A line with a lower slope represent a faster moving object. Just don't go as low as a 45-degree diagonal.

I like to depict the passage of time for a moving object as a series of dots, each one marking a unit of time. This is called Proper Time. For a vertical line, the dots are spaced the same as the coordinate time. For lines that are sloping off the vertical, the dots get farther apart. Here is a spacetime diagram to illustrate what I have been talking about:

The dots on each line show one nsec increments of Proper Time and I have shown ten increments for you (in blue) and ten increments for all the green objects but for the last one in red I have shown only one increment because if I had shown ten, the diagram would be ten times larger. If I had shown an object traveling at 99.99%c it would have had its first increment of Proper Time at 70 nsec and 70 feet. At 99.999%c these numbers would be over 223 and at 99.9999%c they would be over 707. As the speed of an object approaches that of light, the first increment approaches infinity so we don't want to make a drawing of an object traveling too close to it.

I have shown a photon as the thin red line extending out from the red object. Proper Time does not apply to it.

EDIT: The green objects are traveling at successively faster speeds. The first one, next to the blue one is traveling at 0.1c. You can tell that because in 10 nsecs, when it crosses the thin black line, it has traveled 1 foot and since light travels at 1 foot per nsec, the green object is traveling at 0.1c.

When the second green line crosses the thin black line, it has traveled 2 feet and so on until we get to the ninth green line which has traveled 9 feet in 10 nsec.

The thick red line is for an object traveling at 99.9% of the speed of light with only one increment of time.

I could have drawn another set of lines going off to the left to represent negative speeds but I think you can get the idea from just the positive ones, correct?

I don't think the equation for calculating the spacing of the dots at different speeds is complex and if you're interested, let me know and I'll show it to you. HINT: it's called gamma.

Any questions?

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I tried to think of a graph in witch the y is time and the x is distance. This line could be at any positive slope.
That is a space-propertime (or Epstein) diagram as shown here:
Note that it is different from the standard space-coordinatetime (or Minkowski) diagrams, where the aging (propertime) is not visible directly as a length. See this for comparison:

So the only relation is the "speed" at witch the line moves on the graph.
Yes, the postulate here is that everything advances in space-propertime at the same rate. Just the direction varies.

There is not much more to say here. We observe nature to behave that way and we invent geometrical interpretations to make it more intuitive.

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Gold Member
Let's say you're sitting in a train that is circling the Earth at near the speed of light, you're windows are blacked out and the only thing you can see is what is inside your train (the moving train is your frame of reference). Everything would appear normal, you would get up, walk around, drink, eat, talk, breath and everything would appear normal, because everything inside the train is inside the same frame of reference, but the stationary people outside on the platform looking in as you whiz past would see something totally different, they would see you barely moving, you would be in super mega slow motion because relative to the people outside the train, you're moving in slow motion.

Assume you are sitting at the table on the train and the people outside were looking in as you whiz past, they'd have to watch the train for 1 week their time, just to see you lift a glass of water to your mouth (example figures) because of the time difference, but you don't notice this... as far as you're concerned it took you 2 seconds to lift the glass and take a sip of water.

Finally, if you looked out the window at the people who are moving at normal speeds, walking driving ect, it would look like they're whizing all over the place at super speeds, again this is just the time difference.
I am not an expert and it is not my intention to jump all over you, but that last statement is the opposite of what would be observed. The people on the platform will be whizzing past from the point of view of the observers on the train, simply because of the relative motion, just as the training is whizzing past from the point of view of the observers on the platform. However, when a person on the train observes somebody on the platform lifting a glass of water they will also have to wait a week to see them complete that action. The time dilation effect is completely symmetrical.
Actually, in this case, MathJakob is correct because you overlooked the beginning of his scenario which I have included in his quote. His train is not inertial, it is circling the Earth so each time the train passes the station, about seven times per second in the Earth frame, the people on the train have experienced very little time compared to the Earth and it is not symmetrical.

If we are talking about an inertial train (actually, it would have to be a spaceship) that passes by Earth one time, then we have a symmetrical situation in which while they are approaching, they each see the others time progressing more rapidly than their own and then when they pass, they see the others time progressing more slowly than their own but when they calculate the time dilation of each other, they both get the same constant answer.

So in the "orbiting" train, there is a momentary approximate symmetry during each pass in which they each see the others time progressing more rapidly than their own while approaching and then after they pass they each see the others time progressing more slowly than their own but if they calculate the constant "time dilation" of the other, it would not be symmetrical but the reciprocal of each other.

Gold Member
I have two identical stationary objects, for both of those objects, time is exactly the same. For every second that passes for one object, one second will pass for the other object. Now if we accelerate one of the objects, even by a small amount, they will have different frames of reference.

This means that for the stationary object, when one second passes, 0.999999999 of a second will pass for the object that is moving (example figures)

So time for the moving object is slower relative to the stationary object. Now if you accelerated the other object to the same speed, they would both have the same frame of reference again. Their times are synced if you will. They are the same.
If I take your expression that "they would both have the same frame of reference again" to mean they are both at rest in the same frame of reference, then their times will not be synced, assuming that you mean what is normally meant by "synced", namely that their clocks are synchronized according to Einstein's convention.

Let me show you what I mean with some spacetime diagrams. The first one shows the original common rest frame for the two objects (prior to time zero, both objects are at the same place so only the blue one shows up but realize that the red one is also present there):

The dots represent intervals of one second for each object. They have both reset their clocks to zero at the coordinate time of zero. As soon as the blue object accelerates away from the red object, his clock slows down and the two clocks are out of sync since they are ticking at different rates. Now after the red object accelerates to the same speed as the blue object, their two clocks tick at the same rate but they remain out of sync. For example, at the coordinate time of 13 seconds, the red clock is at 12 seconds but the blue one is a little less the 10.5 seconds. You will have to count the dots from time zero to see this.

Now we transform the original spacetime diagram to the mutual rest frame at the end of the scenario:

At the coordinate time of 13, red's clock is at 11 and blue's is at 13. They are obviously ticking at the same rate but they are out of sync.

There is a frame that we can depict the scenario in for which the two clocks will remain in sync from start to finish and that is one in which the two objects are always traveling at the same speed of 0.333c but they change their directions when they accelerate:

At the coordinate time of 13, they both are at just over 12.

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yuiop
Actually, in this case, MathJakob is correct because you overlooked the beginning of his scenario which I have included in his quote. His train is not inertial, it is circling the Earth so each time the train passes the station, about seven times per second in the Earth frame, the people on the train have experienced very little time compared to the Earth and it is not symmetrical.
Yes, I stand corrected. In this none inertial example, the symmetry is broken and the passengers on the train will have aged less than the people on the platform when they eventually come to a stop and get of the train. For measurements made during successive 'orbits' there is a stroboscopic effect and the passengers on the train see the activities of the people on the platform like a sped up movie. This is probably not the best example to introduce classic time dilation in relativity.

If we are talking about an inertial train (actually, it would have to be a spaceship) that passes by Earth one time, then we have a symmetrical situation in which while they are approaching, they each see the others time progressing more rapidly than their own and then when they pass, they see the others time progressing more slowly than their own but when they calculate the time dilation of each other, they both get the same constant answer.
Agree. There is of course a point just before they pass each other when they see each others time progressing at the same rate. At the instant they pass each other, they both see each others time running slower by a factor of gamma. By "see" we literally mean the optical effect due to red or blue shift.
So in the "orbiting" train, there is a momentary approximate symmetry during each pass in which they each see the others time progressing more rapidly than their own while approaching and then after they pass they each see the others time progressing more slowly than their own but if they calculate the constant "time dilation" of the other, it would not be symmetrical but the reciprocal of each other.
That depends on how they calculate the time dilation or which synchronisation convention they use. If the clocks on the train are synchronised using the Einstein convention (so the speed of light is isometric on the train) then measurements made during a very short interval on a single pass of the platform, will show symmetrical time dilation. If the clocks on the train are synchronised in such a way that simultaneous events measured by people on the platform are also simultaneous according to passengers on the train, then the momentary relationship between time rate of clocks on the train and time rate of clocks on the platform will be as you say, reciprocal. This form of synchronisation is in better agreement with the stroboscopic optical effect they see and the end result of comparing clocks and ages when they get off the train.

yuiop
No, you can't travel on a line at any positive slope, the slope must be greater than 1 (and it can also be more negative than -1). It just can't be 1 because that is the speed of light and slopes smaller than that would be faster than light.
I think what adimantium had in mine, was something like the example given by Brian Green that an object that is stationary relative to us, has motion entirely in the 'time' direction (in 4 space) and a particle that is traveling at the speed of light has motion entirely through space. In the 4 space chart, for any positive proper time interval and any positive spatial interval, positive slopes of greater than 1 are allowed.

In the above diagram, the blue vector represents an object that is stationary (in 2 space) moves only in the proper time direction. The black vectors represent objects with timelike velocities that have 4 velocity components in both the proper time direction and the space direction. The red vector represents a photon that travels entirely in the space direction and has a zero proper time component.

The stationary object has zero velocity in 3 space, i.e. dx/dt = 0 but its velocity in 4 space is dtau/dt = c. The magnitude of the 4 velocity of all the particles is in fact also c, and it is only the direction (through proper time or space) that changes.

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