Koveras00
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Could anybody explain the theory behind time dilation and how exactly time dilation works?
Originally posted by Koveras00
Could anybody explain the theory behind time dilation and how exactly time dilation works?
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.
Originally 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?
Originally posted by chroot
jackle,
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
Originally posted by Stranger
Another easy meathod to visualize time dilation can be this...
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...
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.
Originally posted by LURCH
Regarding the second part of your question; I'm afraid I have no idea why the 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.
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
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), ...
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
first explain what is time for you 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?
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
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
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
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?
Originally posted by jackle
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
Originally posted by LURCH
That is what makes Special Relativity so special; there is no difference between the two situations you described!
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).
Originally 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?
Originally posted by pmb
=================
------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
Originally posted by jackle
no, the blue banana will travel rather slowly compared to light at speed b relative to observer 1 and b* relative to observer 2 and will have the classical equation for adding velocities:
b is not equal to b*
classically: b=sqrt(b*^2+v^2) because you are adding velocities which are at right angels
The bananna will seem faster to first observer because when the second observer threw the bananna it had his own speed added to it (speed v).
The second observer who threw the banana will not notice the effect of speed v on the bananna at all because he himself is traveling at speed v.
Originally posted by Koveras00
What if the light is swapped with a moving object moving at a much more slower speed? Will there still be time dilation?? And will the speed of the moving object be the same to the two observers?
Originally posted by mich
Thank you for replying 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
Originally posted by pmb
They're the exact same phenomena. The *reason* whey the wavelength changes is *because* the period decreases. Note that
c = Period/Wavelength = T/L --> L =T/c
Note that Frequency is the reciprocal of period = f = 1/T
L = 1/fc
So you can say that the distance the wave travels from source to observer decreases as the source is approaching and the distance becomes longer when the source is receding. The distance the wave travels happens during the period equal to the reciprocal of the frequency, Therefore the wavelentgh increases when the source is approaching and decreases as the source is receding.
Originally posted by LURCH
Upon further reflection, I realize that the post in which I had described Brian Green's analogy of time dilation was in PF 2.0, so I beg everyone's patients as I repeat myself:
Imagine a dry lake bed. On this lake bed are parallel lines drawn north-to-south. These lines are exactly one mile part. A car driving exactly 60 mph directly east cross this lake bed will across one line every minute. However, if the car were to travel northeast at a 45o angle, (maintaining a speed of 60 mph) it would take two minutes to get from one line to the next. Although the total speed of the vehicle has remained constant, half of that speed is now be expended to achieve northward progress, leaving only half to achieve eastward progress.
It is Mr. Green's contention that we can think of all objects in the universe as having a total velocity of c. Under normal circumstances, most of this velocity is expended in progress through time. However, any motion in any of the three other directions is subtracted from forward progress through time. This is time dilation.
If the car were to travel straight north, it would never cross the next line eastward on lake bed. All of its total velocity (of 60 mph) would be devoted to traveling northward, leaving none to achieve eastward progress. In the same way, any object that devotes its total velocity of c to progress through any of the three spatial dimensions will cease to make progress through the dimension of time. It will never reach the "next moment" in time, because it will be traveling parallel to it.
Originally posted by pmb
Hi mich
No. That is incorrect. The frequency change as well as the wavelength change has nothing to do with the medium. It has to do with the shorter/longer distances the signal has to travel.
."
That also is not true. Light can be thought of as being composed of particles - photons. These photons do *not* change speed. The speed of the photon is indepenant of the speed of the source.
However their energy (does* change with the speed of its source. And the wavelength associated with that energy also changes with the motion of the source.
re - "During that year, the observer decides to speed towards the star (which exists no more). The light will be blue shifted...but what caused the shift?"
The shift is due to the shorted distance the wave has to travel.
Pete
Originally posted by pmb
I made a new web page for this. See
http://www.geocities.com/physics_world/light_clock.htm
There is a derivation in this page which is pretty easy to follow.
Pete
My problem is that the sketch is 2 dimentional, and misses the dimention of depth. The observer who views the moving clock can only see the light which comes towards him/her.
As the moving clock passes by the observer, the light coming towards him/her, will first be blue shifted, as the light has less and less distance to travel. After the
clock passes the observer, the light will be redshifted making the event of the light returning to the ground as being longer than the event of the light hitting the ceiling, making the triangle to be not an isosceles one=
Originally posted by pmb
Hi mich
That page was not intended to explain doppler. Its just a nice easy way to determine what the proper time is compared with the "observer's time".
The frequency of the light does not alter the geometry of th diagram. The triangle is isosceles and I can't see how you concluded that it wasn't based on frequency.
Pete
But it seems that the doppler effect does have something to do with this, since the observer cannot see the proper time due to the light shifts...during red shift the event "seems" slower than proper time, during blueshift the event "seems" faster than proper time.This is the what we could call the observer's time, it seems.
Originally posted by pmb
I see the problem now. You're confusing what is "seen" to what is "measured". They are very different things. It's incorrect to base what you see on what actually is. The rate a which a clock ticks depends only on the speed of the clock. It is not running fast as it approaches us and then runs slow as it moves away. IT may look that way but that is a doppler effect. Doppler is not a time dilation phenomena per se. Even if the moving clock ran at the same rate as the stationary clock there would still be a doppler shift.
Think of it like this. We can put light detectors along the path of the clock. At each detector we have a clock and a recording device. When the light passes it then it notes the time and records the event. All of the clocks are synchronized in the rest frame. We then later gather up all the data and deduce the rate at which the clock ticked. It's kind of a bookkeeping system. From the books we analyze the data and then we'd conclude that the clock ran slower!
Pete [/B]