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- Thread starter Darius Macab
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Thanks

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mathman

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At least that's how I'll look at it until shown to be incorrect.:uhh:

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but c is a constant therefore time must be different in inertial frames at different speeds.

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Pengwuino

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Indeed.

Also, the explanation to 'why' is pretty complicated and can't be summed up in 1 line. Experimentally it has been shown that the speed of light for any observer moving at any velocity relative to anything else will be constant.

The question relates to how one measures distance and velocities and times. For example, if an observer at rest sees a train moving 100mph to the right and another observer is driving by the train going 60mph, the person in the car will only measure the train going at 40mph. So typically, observers measure objects traveling at different speeds depending on their own velocities. Experimentally, however, it has been shown that the speed of light is measured as the same velocity for all observers, regardless of their velocity!

Since any velocity measurement is simply a change in distance during some change in time, for two observers to measure something at the same velocity yet the observers themselves are moving at different velocities during those measurements, something has to be different with how they measure time and distance. The details of that, given correctly by the Lorentz transformations, tell us that people will measure time changing differently and specifically, the observer moving relative to the guy initially at rest will experience a slower time change.

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Similarly seeing as space and time are two aspects of the one spacetime, then if you are not moving time is going as fast as it can. But once you start moving in space then the time component slows down because some of the time component of spacetime is being used up by travelling through space.

So if you travel at the speed of light then you are travelling through space as fast as you can so you are using all the time component of your spacetime motion on travelling through space. Therefore time stands still.

This is how I understood what was being said, anyway.

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the observer moving relative to the guy initially at rest will experience a slower time change.

Hm ... I wonder if that's really the right way to say it. As I understand it, the guy moving will not EXPERIENCE a slower time change but he will have UNDERGONE a slower time change even though not experiencing it. I think this is more than semantic nitpicking.

The guy who travels off at 99.99% of the speed of light (forget about getting squashed by initial G force in this thought experiment) and travels in a huge arc back to the starting point will feel that he has been traveling through time at a perfectly normal rate, but when he gets back he'll see that a clock on the ground that was synchronized with his to start with is now far into the future of his clock even though he EXPERIENCED time as normal.

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Pengwuino

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Hm ... I wonder if that's really the right way to say it. As I understand it, the guy moving will not EXPERIENCE a slower time change but he will have UNDERGONE a slower time change even though not experiencing it. I think this is more than semantic nitpicking.

The guy who travels off at 99.99% of the speed of light (forget about getting squashed by initial G force in this thought experiment) and travels in a huge arc back to the starting point will feel that he has been traveling through time at a perfectly normal rate, but when he gets back he'll see that a clock on the ground that was synchronized with his to start with is now far into the future of his clock even though he EXPERIENCED time as normal.

Yes that was a poor way of stating it. He will experience time like usual, but upon returning to compare, his clock will have advanced much less than the original one.

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If the stopwatch you are using to measure out that 100s is on your body when you travel that fast, 100s.

If the stopwatch is not moving at all,

[itex]\gamma = \frac{1}{\sqrt{1-\beta^2}} [/itex]

with

[itex]\beta = \frac{v}{c} [/itex]

and

[itex]\Delta t = \frac{\tau}{\gamma}[/itex]

Hence,

[itex]t = 14.1067 s [/itex]

Now, if you went 0.9999c,

[itex]t = 1.41418 s [/itex]

And if you were only going 0.5c

[itex]t = 96.8246 s [/itex]

So, notice that you really have to be traveling VERY close to the speed of light to see an effect.

----EDIT----

This is the time you would experience out of the stationary frame's 100s. If you want to see how long the other would experience just multiply 100s by gamma.

Secondly, I fixed my numbers, the were a bit off.

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Would flying around the sun at 1/2 the speed of light in a gravitational field that was immensely stronger than that on earth cause the clock to differ in the same manner as they would if there was no gravity but you were going very close to the speed of light? I know the answer is yes, just curious if anyone knows how the gravitational effect changes the difference in the clocks in comparison to the speed concept? Then we will know which way might be achievable before the other. Or is the combo the best approach? Sending yourself into the future might be useful if you were gravely sick for instance, or just currious :)

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Then why does one person age more than the other?

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Then why does one person age more than the other?

Have you read the thread? Read posts 8 & 10

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Relative to you, the photon clock isn't moving, so the time it takes for the photon to go up and down is simply 2d/c, where c is the speed of light. However, I am outside of the train looking in. I see the photon moving to the side as well as up and down, because the train is moving at speed v. Since the speed of light is a constant, I measure the time for the photon to go up and down as a value greater than 2d/c because the distance travelled by the photon is further.

Thus for me, a greater time has elapsed than for you, even though we have both seen the same event occurring. And so we conclude that the faster you move, the more time slows down.

- #19

ghwellsjr

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Furthermore, even though an observer you left behind sees and measures time for you to be going slower, you see and measure his time to be going slower by the same amount, even though he never accelerated. You are both in the same boat as far as "time slowing down" or as far as "approaching c" goes. To you, he appears to be the one who is traveling at some high rate of speed and therefore "approaching c" and to him, you are the one who is moving at the same high rate of speed and therefore "approaching c".

And to top it off, you don't even have to accelerate at all to have your "time slowing down" or "approaching c". All you have to do is transform a frame in which you are at rest to one in which you are "approaching c" as close as you want and in that frame your time will be slowed down as much as you want. Then, you can see what happens if you actually accelerate in the opposite direction approaching c and then, according to that frame, your time will be speeding up.

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Hm ... I wonder if that's really the right way to say it. As I understand it, the guy moving will not EXPERIENCE a slower time change but he will have UNDERGONE a slower time change even though not experiencing it. I think this is more than semantic nitpicking.

The guy who travels off at 99.99% of the speed of light (forget about getting squashed by initial G force in this thought experiment) andtravels in a huge arc back to the starting pointwill feel that he has been traveling through time at a perfectly normal rate, but when he gets back he'll see that a clock on the ground that was synchronized with his to start with is now far into the future of his clock even though he EXPERIENCED time as normal.

Hi,

I am out of touch in physics for more than a decade, so please excuse my obsolete and corrupt memory/understanding of the concepts...

As per my understanding velocity is a vector quantity, and if the traveler indeed comes back to the starting point - while

The time dilation effect should come into picture, when the person travelling at 99.99%c, tries to read the stationary clock - since that information from stationary clock can at best be sent at c... so the 1 second ticks sent by stationary clock to the person travelling at 99.99% c, will not be 1 second ticks read as per the clock with the travelling man.... and vice versa ( if the stationary person tries to read the clock travelling at 99.99%c)

So, if the traveller continue travelling at 99.99%c, in the direction away from the stationary clock - the time ticks sent by stationary clock when read on traveller's clock will appear dilated - relatively.

However, I am not sure what happens, when the traveller takes a U-turn and approaches the stationary clock.... and if it must come to rest to read the clock, there would be a process of deceleration also - and it would no more be non-accelerating frames of observation....the traveller must read the stationary clock, while maintaining the constant speed.

- #21

ghwellsjr

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No, it doesn't matter his direction, his speed is what counts relative to some frame, in this case the stationary clock. So the traveling clock always experiences the same time dilation throughout the entire journey.Hi,

I am out of touch in physics for more than a decade, so please excuse my obsolete and corrupt memory/understanding of the concepts...

As per my understanding velocity is a vector quantity, and if the traveler indeed comes back to the starting point - whiletravelling in the arc- the velocity vector relative to the stationary clock will not always be 0.9999c, half of the journey time should dilate, and the other half, it would be other wise... so when he does comes back and stops, there should not be any time difference.

You are correct, each observer will read the other one's clock as ticking slower by exactly the same amount but here we are not talking about time dilation, we are talking about Relativistic Doppler which is what each observer actually sees of the other one's clock. The amount of slowdown that they see is not the amount of time dilation.The time dilation effect should come into picture, when the person travelling at 99.99%c, tries to read the stationary clock - since that information from stationary clock can at best be sent at c... so the 1 second ticks sent by stationary clock to the person travelling at 99.99% c, will not be 1 second ticks read as per the clock with the travelling man.... and vice versa ( if the stationary person tries to read the clock travelling at 99.99%c)

What happens when the traveler makes the U-turn, is that he immediately sees the stationary clock speed up. This the Relativistic Doppler in the other direction, exactly the inverse factor, but time dilation is still going on.So, if the traveller continue travelling at 99.99%c, in the direction away from the stationary clock - the time ticks sent by stationary clock when read on traveller's clock will appear dilated - relatively.

However, I am not sure what happens, when the traveller takes a U-turn and approaches the stationary clock.... and if it must come to rest to read the clock, there would be a process of deceleration also - and it would no more be non-accelerating frames of observation....the traveller must read the stationary clock, while maintaining the constant speed.

The stationary observer does not see this happen in the traveler's clock until a long time later. It is this imbalance in the times that each one sees of the other's clock running slow then fast that accounts for the difference in the clocks elapsed time when they rejoin.

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Because the frequency of an oscillator is directly related to the speed at which the oscillator is moving. This includes the atoms that constitute the biological oscillators of an organism. So, the faster you move, the less you age.Then why does one person age more than the other?

While the mainstream geometrical interpretation (Minkowski) of SR is quantitatively correct, there's no detailed mechanistic understanding of differential aging yet -- though the quantitatively equivalent Lorentz ether theory is a step in that direction.

As others have noted, nobody will feel like they're aging any differently as their speed increases or decreases, and their own clocks won't appear to them to be slowing down or speeding up. But they will feel something if their speed increases or decreases, and it's during these intervals of changing speeds that the changes in biological oscillators, as well as the, say, crystal oscillator that's the basis of a reference clock, are occuring.

But this shouldn't be confused with what's actually determining the accumulated difference in age between, say, a person at rest on earth and a person moving at an average of, say, .5 c in a spaceship. This accumulated difference is solely determined by the duration of intervals wrt which there are differences in speed between the two.

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- #23

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While the mainstream geometrical interpretation (Minkowski) of SR is quantitatively correct, there's no detailed mechanistic understanding of differential aging yet -- though the quantitatively equivalent Lorentz ether theory is a step in that direction.

That's what I was thinking as I read the posts above....nothing wrong with the above explanations as ways of thinking about it. I like Brian Greene's explanation and have used it myself, but nobody really knows WHY the speed of light is constant while distance and time vary among inertial frames.

It's akin to asking WHY the electron has the mass...or the charge.... it does....or why we have the four fundamental forces we observe....we know they are there but not WHY.

good discusssion overall.

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ghwellsjr

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So you're saying that the person that accelerated to 0.5c is accumulating less age than the person at rest because of the time duration that he spends at that speed? And this is better explained by LET than SR? And that if he increases his speed, his clock slows down and if he decreases his speed, his clock speeds up? It sure sounds like you are promoting an absolute state of rest. Can you explain further, because I'm afraid that Garry will get a wrong impression?Because the frequency of an oscillator is directly related to the speed at which the oscillator is moving. This includes the atoms that constitute the biological oscillators of an organism. So, the faster you move, the less you age.Then why does one person age more than the other?

While the mainstream geometrical interpretation (Minkowski) of SR is quantitatively correct, there's no detailed mechanistic understanding of differential aging yet -- though the quantitatively equivalent Lorentz ether theory is a step in that direction.

As others have noted, nobody will feel like they're aging any differently as their speed increases or decreases, and their own clocks won't appear to them to be slowing down or speeding up. But they will feel something if their speed increases or decreases, and it's during these intervals of changing speeds that the changes in biological oscillators, as well as the, say, crystal oscillator that's the basis of a reference clock, are occuring.

But this shouldn't be confused with what's actually determining the accumulated difference in age between, say, a person at rest on earth and a person moving at an average of, say, .5 c in a spaceship. This accumulated difference is solely determined by the duration of intervals wrt which there are differences in speed between the two.

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The difference in time (or age) that the (earthbound and travelling) biological oscillators accumulate during an interval is due to their difference in speed during that interval. A traveller moving at an average of, say, .5c for a roundtrip interval of, say, 30 years (earth time) will have aged noticeably less than the earthbound person (by an amount given by the Lorentz transformation).So you're saying that the person that accelerated to 0.5c is accumulating less age than the person at rest because of the time duration that he spends at that speed?

Yes.And that if he increases his speed, his clock slows down and if he decreases his speed, his clock speeds up?

Not necessarily. They're quantitatively equivalent. But to even begin to have a mechanistic understanding of why an oscillator's frequency decreases as the oscillator's speed increases, and vice versa, then you'd have to have the oscillator interacting with something.And this is better explained by LET than SR?

As I mentioned, LET is a step in that direction. But it's certainly not a detailed account of what's happening with oscillator's as they move about. And SR is even less detailed in that regard.

I don't know what that might refer to. Relativity says that the laws of physics don't depend on states of motion, and, so far, that seems to be the case.It sure sounds like you are promoting an absolute state of rest.

I do think that we can reasonably infer the existence of a wave/particle reality underlying our observations, and that while SR obviates that view it certainly doesn't rule it out.

What I suggested is that the frequency of an oscillator is directly related to the speed at which the oscillator is moving, and that there's currently no detailed mechanical explanation of how a change in the speed of an oscillator produces a change in the frequency of the oscillator.Can you explain further, because I'm afraid that Garry will get a wrong impression?

What other impression might he get from anything I've said?

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