Traveling near the speed of light

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So the theory does apply, since if I were to suppose an observer were moving at c, the theory will tell me what this means? For example, as v approaches c, γ approaches ∞, so if we were to consider the total energy of an observer of mass m with speed c,

E = γmc2 = ∞

Thus, in order to be moving at c, the observer must have a total amount of energy that is infinite, but it is impossible to transfer an infinite amount of energy to any massive object since this would contradict the Law of Conservation of Energy.
If the observer is were a photon...?
 
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I intended the previous post is a pure speculative question or light joke.[Add: not a joke about "light]
Now this is not.
An observer can't be a photon. An "observer" has to have a rest frame, and a photon doesn't.
1. A photon doesn't have a rest frame, because as pictured in the space time diagram, it is geometrically impossible to boost so that light ray is vertical?
What "problem" are you trying to resolve? Why does Red choose those particular values to send to Blue and Green? What is Red trying to accomplish?
2. An observer has to have a rest frame. But the observer can choose any moving/rest frame to make the problem easier? Is that so?
Consider this picture.
Red DOES have a rest frame, but to calculate problem such as the picture below, we can choose the picture at the right side so we can calculate simultaneity events for Blue and Green easily.
Synch-Small.jpg
 
No. The theory says that only things with zero rest mass (like light) can move at c; an "observer" (meaning something like you or me or a spaceship or a measuring device) can't. That's because any "observer" must have nonzero rest mass.
I think I may have used poor communication here. I understand very well that the theory tells you it is not possible for any object with a rest mass to move at c. I think my question,

So the theory does apply
arrow-10x10.png
, since if I were to suppose an observer were moving at c, the theory will tell me what this means?
was a little ambiguous. So the theory applies to light because it correctly predicts the behavior of light and also tells you that objects with rest mass can not move like light?

Have you looked at any basic SR textbooks? If you haven't, you should. This is all very basic SR.
There is a chapter in my physics textbook on SR that I have read. I have not read any textbooks that specialize in SR. I thought I had a basic understanding of the theory.

here is no such thing; a light beam moves at c in every reference frame, so there can never be a frame in which it at rest, and "the reference frame of a light beam" assumes that the light beam is at rest in the frame.

Again, if you haven't looked at an SR textbook, you should. This is very basic SR. (There is also a Physics Forums FAQ explaining that there is no such thing as the rest frame of a photon.)
I see. I was unaware of this and clearly my knowledge is limited. So I understand that my other example that considered the frame of the light beam is invalid. So then this means SR can not model light in this way?

No. The ship is moving relative to Earth and the beacon, so the difference in light travel times from Earth to the ship, vs. from the beacon to the ship, is not constant.
Is it not constant because light actually travels at constant speed in any inertial reference frame?


The difference in clock readings between the beacon and Earth, in the ship frame, is due to relativity of simultaneity.
My understanding of relativity of simultaneity is also basic and perhaps it is not adequate enough as I am unsure how relativity of simultaneity would justify the fact that the beacon's clock reads different than the Earth's clock.

Would you mind recommending me a few books?
 
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A photon doesn't have a rest frame, because as pictured in the space time diagram, it is geometrically impossible to boost so that light ray is vertical?
That's not the reason, but it's a way of visualizing it, yes. The reason a photon doesn't have a rest frame is the law that photons move at ##c## in all reference frames, so there can't be a frame in which a photon is at rest. In terms of the spacetime diagram, a Lorentz transformation leaves the lightlike lines (45 degree lines) invariant; it doesn't change their slope at all, so yes, the diagram correctly represents the underlying law that photons move at ##c## in all frames.

An observer has to have a rest frame. But the observer can choose any moving/rest frame to make the problem easier?
Yes. But he can't choose a frame in which a photon is at rest, because there is no such frame.
 
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So the theory applies to light because it correctly predicts the behavior of light and also tells you that objects with rest mass can not move like light?
Yes.

this means SR can not model light in this way?
Yes. The way SR models light, or anything with zero rest mass, works differently from the way it models things with nonzero rest mass. There are still many things in common between the two ways of modeling, but the differences are fundamental and important.

Is it not constant because light actually travels at constant speed in any inertial reference frame?
That's part of it; the other part is that the distances are changing because the ship is moving relative to Earth and the beacon.

I am unsure how relativity of simultaneity would justify the fact that the beacon's clock reads different than the Earth's clock.
The beacon's clock and the Earth's clock are synchronized in the Earth-beacon rest frame--that is, a given reading on either clock is simultaneous, in that frame, with the same reading on the other clock. But if that's true in the Earth-beacon frame, it won't be true in any other frame, by relativity of simultaneity; events that are simultaneous in one frame are not simultaneous in any other frame. So in any other frame, a given reading on, say, the Earth clock will not be simultaneous with the same reading on the beacon clock; instead, it will be simultaneous with some other reading on the beacon clock. To find out which readings on the two clocks will be simultaneous, you need to do the math to transform between the frames; that's how I derived the numbers in my earlier post.

Would you mind recommending me a few books?
Taylor & Wheeler's Spacetime Physics is a good text. A good online reference is here:

http://lightandmatter.com/sr/
 
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I intended the previous post is a pure speculative question or light joke.[Add: not a joke about "light]2. An observer has to have a rest frame. But the observer can choose any moving/rest frame to make the problem easier? Is that so?
Yes. But [..]
THANKS. It helps me clear many problems! I tough this is a mathematic quiz. Like "Solve this problem with a WL moves at certain V" And we are not supposed to boost the diagram. But we can freely boost the diagram anyway we like to solve the problem.
That's why I don't didn't understand [add: CLEARLY] your post earlier.
What "problem" are you trying to resolve? Why does Red choose those particular values to send to Blue and Green? What is Red trying to accomplish?
Now I understand it. Crystal clear.
 
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One more question:
To solve the numbers (coordinate of events, wordlines direction) we can boost the ST diagram in any V we want.
But to imagine what the nature would look like, we should boost it again, so we are at the rest frame. Is that so?
 
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We DON'T travel!. It's the universe that travel approaches the speed of light. It's the universe that won't age. We still feel 1 second is 1 second.
I remember an Einstein joke (or anectode?)
Once on a train in US (Einstein had already been in US), a student recognized him. And the student ask, "I'm sorry professor, when will New York stop at the train?"
How do you beat gravity force? Jump out of the window, and the floor (and the earth, the moon, everything) will move toward you accelerated at 9.8ms. As long as you stay on the room, you are actually travel and accelerated at 9.8 m/s^2
BUT DON'T PROVE IT!:smile:
Do you mean that if someone were to live on Earth 85 years, (s)he would also live 85 years on the spaceship (assuming the same life style on Earth and spaceship) according to the spaceship's clock while (s)he is traveling at a speed close to the speed of light, regardless of the difference between the Earth's and spaceship's clocks?
 
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To solve the numbers (coordinate of events, wordlines direction) we can boost the ST diagram in any V we want.
Yes.

to imagine what the nature would look like, we should boost it again, so we are at the rest frame.
I'm not sure I understand. How is this different from boosting the ST diagram in any V we want?
 
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Do you mean that if someone were to live on Earth 85 years, (s)he would also live 85 years on the spaceship (assuming the same life style on Earth and spaceship) according to the spaceship's clock
Yes.
 
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Yes.



I'm not sure I understand. How is this different from boosting the ST diagram in any V we want?
Come on, you know better
.
ST-01.JPG

To calculate simultaneity of E1 and E2, we (us) have to use the big picture. What the world really look like? We (us) use the right bottom picture. Here, we see that blue and red speed actually differ more than we see in the big picture.
 
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Come on, you know better
No, I don't. If I understood what you were trying to say, I wouldn't have asked the question I asked.

What the world really look like?
This is a meaningless question; no particular frame tells you what the world "really" looks like. "Really" is not a scientific word.

Here, we see that blue and red speed actually differ more than we see in the big picture.
No, that's not what we see. First of all, as above, no particular frame tells you what is "actually" the case. "Actually" is not a scientific word.

Second, the relative velocity of blue and red is an invariant; it doesn't depend on which frame you choose. If you are getting the result that the relative velocity of blue and red is different in different frames, you are doing something wrong.
 
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This is a meaningless question; no particular frame tells you what the world "really" looks like. "Really" is not a scientific word.
No, that's not what we see. First of all, as above, no particular frame tells you what is "actually" the case. "Actually" is not a scientific word.
Good point. Now I know what scientific method is.
Second, the relative velocity of blue and red is an invariant; it doesn't depend on which frame you choose. If you are getting the result that the relative velocity of blue and red is different in different frames, you are doing something wrong.
Just the angle look different.
It's the ##\frac{u+v}{1+uv}##
 

HallsofIvy

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No. That is the whole point of relativity- all photons move at speed "c" relative to all reference frames. To a person "A" traveling toward another star at a large percentage of the speed of light, relative to person "B" on earth, the distance to that star is shorter that it is to "B". To person "B", "A" is traveling the greater distance but his time scale is dilated. So "distance divided by time" works out the same to each.
 
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So what does that mean? I still don't understand what point you're trying to make.
Perhaps I can't express myself clearly.
Supposed you are green in this diagram.
ST-01.JPG

And you want to know if event E1 and E2 are simultaneous, so you boost the ST diagram so that the maroon world line are at rest. Then you'll find that E1 and E2 is simultaneous.
But to see the diagram with naked eye, it seems that red and blue WL is very close to green. Actually in the real world, the angle of red and blue is bigger if you put green at rest. See the yellow rectangle.
And IMHO, twins paradox is no more mistery than barn paradox and train/platform experiment. I'm new here. But in the past three months I think I can grasp basic SR. Still so much to learn tough.
 
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Actually in the real world, the angle of red and blue is bigger if you put green at rest.
What do you mean by "in the real world"? Calculate the relative velocity of red and blue in any frame; it will be the same. That means that, "in the real world", the relative velocity of red and blue is the same for all observers. The fact that the angle appears different on different spacetime diagrams is a fact about the diagrams, not about red and blue or anything "in the real world" (meaning, the objects and motions that the diagrams represent).
 
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[..]That means that, "in the real world", the relative velocity of red and blue is the same for all observers.
GOOD POINT!. I'll never forget that.
The fact that the angle appears different on different spacetime diagrams is a fact about the diagrams, not about red and blue or anything "in the real world" (meaning, the objects and motions that the diagrams represent).
Yes, I do understand that. Like I said, I can't express myself clearly.
Consider this.
VBlue wrt Red is 0.98c
VGreen wrt Blue is 0.6c
For Red, VGreen is 0.995c. Just slighty 0.015c than VBlue. And if you boost the diagram where Blue is at rest, than you'll see that the different/angle between VGreen and VBlue is 0.6c. I know, I know, we have to account for relative velocity addition.
It's like that we say.
"No, it's not Blue (660 THz) with V=0, it's Red (440 THz) with V = 0.2c"
Our eyes see it as Blue, our brain calculates it as Red (should compare it with Franhover lines, otherwise, you won't know if it's blue-shifted).
And in my opinion I think Twins Paradox is no more mystery than barn paradox or train experiment.
Perhaps the word "Paradox" came out when Relativiy was at its infancy, and the most natural phenomena such as time dilation for one of the twins was considered "amazing".
PeterDonis, I have a problem with Train Experiment, perhaps you might want to take a look at it in: https://www.physicsforums.com/threads/train-experiment-problem.825514/
Thanks
 

DrGreg

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Perhaps the word "Paradox" came out when Relativiy was at its infancy, and the most natural phenomena such as time dilation for one of the twins was considered "amazing".
Paradoxically, the word "paradox" has (amongst others) two meanings that are almost opposites:
  1. an argument that comes to a false conclusion (e.g.contradicts itself), via steps that appear, at first, to be valid
  2. an argument whose conclusion may appear, at first, to be false, but is actually true
The "twins paradox", like some other paradoxes in maths and physics, is a paradox of the second type.

In this post I used "paradoxically" with a third meaning, "having apparently contradictory characteristics".
 

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