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Could anybody explain the theory behind time dilation and how exactly time dilation works?

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Could anybody explain the theory behind time dilation and how exactly time dilation works?

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ahrkron

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You should probably give a try to Tom's

Physics Napster.

Post number 13 in the first page of the thread has many good links about relativity.

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Consider yourself at a launch pad, wherefrom a rocket is launched vertically at constant speed v. The rocket sends a monochromatic laser signal back to the pad. A counter on the ground collects the number of wavelengths sent out by the craft on its away trip, as it does with the return trip at the same velocity. It is found, with c being the constant velocity of light in all inertial frames, that the wavelength of the light signal for the away trip exceeds the wavelength of the standard laser signal exceeds the wavelength for the return trip. Think of the observed laser wave train (representing a "standard meter") from the spaceship being dilated as it moves out, and contracted as it returns. Time is measured as [del]t=[lamb]/c, where [lamb] is wavelength.

Other variations of the trip include moving horizontally across the point of observation, or diagonal trajectories. The Pythagorean theorem may be used to help derive these special relativistic transformations of length, time, mass, velocity and energy.

Other variations of the trip include moving horizontally across the point of observation, or diagonal trajectories. The Pythagorean theorem may be used to help derive these special relativistic transformations of length, time, mass, velocity and energy.

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Imagine that you are looking out from the porthole of your spaceship into another.....the two ships are passing each other with a uniform velocity close to the speed of light.....as they pass a beam of light on the other ship is sent from its ceiling to its floor...........there it strikes a mirror and is reflected back....you will see the path of light as 'V' while the person in that ship will see it as a straight line.....with some instrument you could clock the time it takes for the beam to traverse teh V shape....by dividing the length and teh path, you obtain the speed of light...

Now while you are doing this....the person in the other ship is doing the same thing.... to his point of view light simply goes up and down along the same line, obviously a shorter distance than along the V path that you observed...when he divides the distance of the straight line path he observes by the time....he gets the speed of light.....because the speed of light is constant for all observors he should get the same answer as yours.....but his light path is shorter...how can the results be the same.....there is only one possible explanation: his clock is slower.... ofcourse the situation is perfectly symmetrical.....

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It works on the basis of the speed of light being constant and most the other physical equations being re-written around that fact. This (at risk of being too simplistic) includes how fast time seems to pass. Time dialation is time taking longer to pass and compensates mathematically for various other things.

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chroot

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I certainly wouldn't say it's incredibly complicated... in fact, I'd say it's very simple. Basic high-school algebra is the only mathematical tool you'll need to compute things in special relativity.

- Warren

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pmb

Consider the two postulates of relativityOriginally posted by Koveras00

Could anybody explain the theory behind time dilation and how exactly time dilation works?

(1) The laws of physics are the same in all inertial frames of referance

(2) The speed of light is independant of the source

Not take a mirror and a light emitter/detector (ED) and place them as follows

=================

------E-D--------

---------------------------------------------------> X

Its easy to see that the distance "L" between the mirror and ED does not change as the apparatus (mirror and ED) move to the right with a given velocity.

Let O be the frame in which the apparatus is at rest. So if a flash of light is emitted at ED which travels to the mirror and bounces back to ED then the time taken as measued in this frame is given by

T = 2L/c

Now consider the same thing from a frame of referance in which the entire apparatus is moving in the X direction with velocity "v" - Call that frame O'. Then (I can't draw that here - it gets messed up) the time taken as determined in the frame O must be greater since the light has to travel a greater distance. So T' > T

Do this out and use the Pathagorean theorem and you'll see that

T' = T/sqrt[1 - (v/c)^2]

Pete

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Originally posted by Loren Booda

Consider yourself at a launch pad, wherefrom a rocket is launched vertically at constant speed v. The rocket sends a monochromatic laser signal back to the pad. A counter on the ground collects the number of wavelengths sent out by the craft on its away trip, as it does with the return trip at the same velocity. It is found, with c being the constant velocity of light in all inertial frames, that the wavelength of the light signal for the away trip exceeds the wavelength of the standard laser signal exceeds the wavelength for the return trip. Think of the observed laser wave train (representing a "standard meter") from the spaceship being dilated as it moves out, and contracted as it returns. Time is measured as [del]t=[lamb]/c, where [lamb] is wavelength.

Can anyone explain how and why the wavelength wil be different during the away and return trip and any explanation behind that the speed of light is constant?

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the simpliest way to understand time dilation is, as you travel near the speed of light, time slows down..

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LURCH

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Regarding the second part of your question; I'm afraid I have no ideaOriginally posted by Koveras00

Can anyone explain how and why the wavelength wil be different during the away and return trip and any explanation behind that the speed of light is constant?

Taking that as a given, I have an annalogy that might help you understand the change in wavelength/frequency of light from a moving source.

Suppose I have a whole bunch of wind-up toy cars. Each car, when placed on the ground, travels 2 meters per second. I place one car on the ground every second, all pointed toward you. Cars will arrive at your location at a rate of 1 per second, and they will be 2 meters apart.

Now suppoose that I begin walking towards you at a rate of 1 meter per second. As I place a car on the ground, it begins traveling towards you. One second later, the car has traveled two meters in your direction, but I have also traveled one meter, so when I set the next car down, it is only one meter behind the previous car. Now cars are arriving at your location at a rate of one every .5 seconds, and they are only one meter appart. Of course, if I walk backwards away from you at the same pace, you will receive one car every 1.5 seconds, and they will be 3 meters appart. So the frequency and distance between them changes because their speed is constant, and mine is not.

- #12

pmb

There is a footnote in another of Einstein's 1905 paper (same issue of journal that he published what we call "special relativity"). The footnote reads

"The constancy of light is contained in Maxwell's Equations"

That means that if you assume that Maxwell's equations hold true then it follows that the constants in those equations are independant of the frame of referance - otherwise you'd be able to speak of a prefered frames of referance. These constants (permitivity (e) and permiability (u) of free space - I think they're related to each other though) determine the speed of light

c = 1/sqrt(eu)

So you really don't need to invoke the constancy of light as a second postulate if you assume Maxwell's equations are valid as they stand.

Pete

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It does depend on the background you come from I think chroot. When I talk about time dialation to a lot of ordinary people they think I have been watching too much Star Trek.Originally posted by chroot

I certainly wouldn't say it's incredibly complicated... in fact, I'd say it's very simple. Basic high-school algebra is the only mathematical tool you'll need to compute things in special relativity.

- Warren

- #14

mich

The problem that I have with this, Stranger, is that in order to "see" the light, its rays need to come towards me first; and let the light's speed coming towards me be constant.Originally posted by Stranger

Imagine that you are looking out from the porthole of your spaceship into another.....the two ships are passing each other with a uniform velocity close to the speed of light.....as they pass a beam of light on the other ship is sent from its ceiling to its floor...........there it strikes a mirror and is reflected back....you will see the path of light as 'V' while the person in that ship will see it as a straight line.....with some instrument you could clock the time it takes for the beam to traverse teh V shape....by dividing the length and teh path, you obtain the speed of light...

Now while you are doing this....the person in the other ship is doing the same thing.... to his point of view light simply goes up and down along the same line, obviously a shorter distance than along the V path that you observed...when he divides the distance of the straight line path he observes by the time....he gets the speed of light.....because the speed of light is constant for all observors he should get the same answer as yours.....but his light path is shorter...how can the results be the same.....there is only one possible explanation: his clock is slower.... ofcourse the situation is perfectly symmetrical.....

When the spaceship is passing by me, the light will first be "blueshifted" as it (spaceship)gets closer, making each ray of light coming towards me as having a shorter path to follow...now this blueshift is not a relavistic shift, but a classical one.Therefore the time of the event for the light hitting the ceiling will be viewed as being shorter than,after the spaceship has passed me, the time of the event for the light travelling back towards the floor,as it will be viewed to be longer, since the light will have been redshifted.

The observed time ought to be equal to both observers, it seems.

mich

- #15

mich

Hello Loren:Originally posted by Loren Booda

Consider yourself at a launch pad, wherefrom a rocket is launched vertically at constant speed v. The rocket sends a monochromatic laser signal back to the pad. A counter on the ground collects the number of wavelengths sent out by the craft on its away trip, as it does with the return trip at the same velocity. It is found, with c being the constant velocity of light in all inertial frames, that the wavelength of the light signal for the away trip exceeds the wavelength of the standard laser signal exceeds the wavelength for the return trip. Think of the observed laser wave train (representing a "standard meter") from the spaceship being dilated as it moves out, and contracted as it returns. Time is measured as [del]t=[lamb]/c, where [lamb] is wavelength.

Other variations of the trip include moving horizontally across the point of observation, or diagonal trajectories. The Pythagorean theorem may be used to help derive these special relativistic transformations of length, time, mass, velocity and energy.

Would you not agree that if light would travel in a medium,

the observation above would also be detected, but without any dilation of time involved?

- #16

mich

Good explanation, Lurch:Originally posted by LURCH

Regarding the second part of your question; I'm afraid I have no ideawhythe speed of light is constant. I'm not sure anyone knows, we just keep measuring it and it allways comes up the same.

Taking that as a given, I have an annalogy that might help you understand the change in wavelength/frequency of light from a moving source.

Suppose I have a whole bunch of wind-up toy cars. Each car, when placed on the ground, travels 2 meters per second. I place one car on the ground every second, all pointed toward you. Cars will arrive at your location at a rate of 1 per second, and they will be 2 meters apart.

Now suppoose that I begin walking towards you at a rate of 1 meter per second. As I place a car on the ground, it begins traveling towards you. One second later, the car has traveled two meters in your direction, but I have also traveled one meter, so when I set the next car down, it is only one meter behind the previous car. Now cars are arriving at your location at a rate of one every .5 seconds, and they are only one meter appart. Of course, if I walk backwards away from you at the same pace, you will receive one car every 1.5 seconds, and they will be 3 meters appart. So the frequency and distance between them changes because their speed is constant, and mine is not.

My question is this, though; How does one explain the change in wavelengths when the observer, not the source changes speed,remaining the speed of light the same?

mich

- #17

mich

The way I see this, Pete, when considering a medium for light, the time involved when the apparatus is moving at velocity v, would be 2L/(c+v)+(c-v)/2= 2L/c (relative to the frame), as well.Originally posted by pmb

Consider the two postulates of relativity

(1) The laws of physics are the same in all inertial frames of referance

(2) The speed of light is independant of the source

Not take a mirror and a light emitter/detector (ED) and place them as follows

=================

------E-D--------

---------------------------------------------------> X

Its easy to see that the distance "L" between the mirror and ED does not change as the apparatus (mirror and ED) move to the right with a given velocity.

Let O be the frame in which the apparatus is at rest. So if a flash of light is emitted at ED which travels to the mirror and bounces back to ED then the time taken as measued in this frame is given by

T = 2L/c

Now consider the same thing from a frame of referance in which the entire apparatus is moving in the X direction with velocity "v" - Call that frame O'. Then (I can't draw that here - it gets messed up) the time taken as determined in the frame O must be greater since the light has to travel a greater distance. So T' > T

Do this out and use the Pathagorean theorem and you'll see that

T' = T/sqrt[1 - (v/c)^2]

Pete

But considering the above postulates, the light must always remain c

nevertheless still leaving the time as being 2L/c.

As for the distance between the emiter and detector changing, I'm not sure why you're saying this.

mich

- #18

pmb

Originally posted by mich

The way I see this, Pete, when considering a medium for light, the time involved when the apparatus is moving at velocity v, would be 2L/(c+v)+(c-v)/2= 2L/c (relative to the frame), ...

I don't understand this equation. Something is wrong with it. Notice that the first term 2L/(c+v) has the dimensions of time yet the qyantity (c-v)/2 has dimensions of distance/time.

Pete

- #19

pmb

The same thing happens with sound. A train comming towards you will have a higher pitched sound then when moving away from you. The reason being can be seen as follows. Instead of a whistle thing in terms of beeps. If the train is stationary and it beeps then the sound will travel to your ear at equal time intervals, the same time intgerval at when they left the train. Now consider what happens when the train is moving. The train passes a pole and it beeps just when it is at that pole. The sound has to move from that pole to your ear. However the train is at a different, closer, location when it emits the second beep. Now that beep has a shorter distance to travel so you hear it sooner than if the train wasn't moving! So you hear the beeps at shorter time intervals. That is you hear the beeps at a different frequency.Originally posted by mich

Good explanation, Lurch:

My question is this, though; How does one explain the change in wavelengths when the observer, not the source changes speed,remaining the speed of light the same?

mich

Same thing with light with one difference - the time between the beeps will be decreased because of time dilation. However the police do not need to take that into account when they uyse this effect when they are at speed traps and using radar to clock the speed of your car! If you get caught try telling him he was the one speeding. You were the one at rest. :-)

This is also the same reason why when you look at a plane or a jet flying in the sky it appears ahead of where the sound is coming from - and that does not mean they are moving faster than sound.

Pete

- #20

dr-dock

first explain what is time for ya and i'll probably tell you about dilationOriginally posted by Koveras00

Could anybody explain the theory behind time dilation and how exactly time dilation works?

- #21

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Mich, I think we need to face observed facts. There are plenty of ways the universe could work mathematically, but only one way that it actually does.

ps. I think you meant:

2L / 0.5[(c+v)+(c-v)]

not:

2L/(c+v) + (c-v)/2

ps. I think you meant:

2L / 0.5[(c+v)+(c-v)]

not:

2L/(c+v) + (c-v)/2

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- #22

LURCH

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That is what makes Special Relativity so special; there is no difference between the two situations you described!Originally posted by mich

My question is this, though; How does one explain the change in wavelengths when the observer, not the source changes speed,remaining the speed of light the same?

mich

My analogy contained one fatal flaw. It used an absolute frame of reference; the floor on which both of us stood and by which the cars propelled themselves. In reality, there is no absolute frame of reference. So saying that the light source is moving toward you is exactly the same as saying that you are moving toward it, and will yield exactly the same observed results.

BTW; Brian Green's book The Elegant Universe contains one of the best analogies I have ever seen to describe time dilation. It seems to me that I have already played it out elsewhere in these Forums, so I will see if I can find that thread (just so I don't take up a bunch of space repeating myself).

- #23

mich

Sorry Pete; I forgot some brakets,as Jackel pointed out.Originally posted by pmb

I don't understand this equation. Something is wrong with it. Notice that the first term 2L/(c+v) has the dimensions of time yet the qyantity (c-v)/2 has dimensions of distance/time.

Pete

T= 2L(distance)/[(c+v)+(c-v)/2](average velocity of light)

- #24

mich

Thank you for replying Pete;Originally posted by pmb

The same thing happens with sound. A train comming towards you will have a higher pitched sound then when moving away from you. The reason being can be seen as follows. Instead of a whistle thing in terms of beeps. If the train is stationary and it beeps then the sound will travel to your ear at equal time intervals, the same time intgerval at when they left the train. Now consider what happens when the train is moving. The train passes a pole and it beeps just when it is at that pole. The sound has to move from that pole to your ear. However the train is at a different, closer, location when it emits the second beep. Now that beep has a shorter distance to travel so you hear it sooner than if the train wasn't moving! So you hear the beeps at shorter time intervals. That is you hear the beeps at a different frequency.

Same thing with light with one difference - the time between the beeps will be decreased because of time dilation. However the police do not need to take that into account when they uyse this effect when they are at speed traps and using radar to clock the speed of your car! If you get caught try telling him he was the one speeding. You were the one at rest. :-)

This is also the same reason why when you look at a plane or a jet flying in the sky it appears ahead of where the sound is coming from - and that does not mean they are moving faster than sound.

Pete

I agree with most of what you wrote, Pete, but,the cause for the shift when the source is moving is due to a change in wavelength while the change of shift which happens when the observer moves, or changes speed is due to a change in the speed of sound relative to the observer. In the case of light, this cannot be the reason.

mich

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The only valid explanation of time dilation comes from understanding the Lorentz transformation equations. This is the first step. If you’d like to see a derivation of these basic equations, go to this link: http://www.everythingimportant.org/relativityOriginally posted by Koveras00

Could anybody explain the theory behind time dilation and how exactly time dilation works?

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