Traveling near the speed of light

[Stephanus]

Yes. Remember that "synchronize" here has a very specific technical meaning: it means "synchronize" in the sense of constructing an inertial frame such that both clocks tick coordinate time in that frame. That requires the clocks to be at rest relative to each other.

You appear to be using "synchronize" in a more general sense, to mean something like "clocks exchanging information about their times and rates". Of course any two clocks can do this using light signals. But that in itself doesn't count as "synchronization" in the technical sense I described above, which is the sense we need if we're talking about inertial frames in SR.

Thanks! It helps me clear the way to understand SR.

 

[Stephanus]

Yes. Remember that "synchronize" here has a very specific technical meaning: it means "synchronize" in the sense of constructing an inertial frame such that both clocks tick coordinate time in that frame. That requires the clocks to be at rest relative to each other.

I'm sorry. I have this scenario in mind.

At event A, Green collect information from Blue. Green knows Blue left/right speed, velocity, distance, etc..
Can't Green at event B after make some calculation send command to Blue left and Blue right.
Left: "Set your clock at so and so"
Right: "Set your clock at so and so"
Green could have done that with pen/paper and watch.

 

[PeterDonis]

Can't Green at event B after make some calculation send command to Blue left and Blue right.

Yes, but the clock readings the Blue clocks are commanded to set will not be consistent with coordinate time in the inertial frame in which the Blue clocks are at rest. (Actually, I'm assuming this would be the case, since otherwise I don't see why you'd be raising the point. But you haven't described the calculation Green does, so we don't know what commands Green would actually send in the scenario you've described.) So this won't be "clock synchronization" in the technical sense I defined.

 

[Stephanus]

Yes, but the clock readings the Blue clocks are commanded to set will not be consistent with coordinate time in the inertial frame in which the Blue clocks are at rest. (Actually, I'm assuming this would be the case, since otherwise I don't see why you'd be raising the point. But you haven't described the calculation Green does, so we don't know what commands Green would actually send in the scenario you've described.) So this won't be "clock synchronization" in the technical sense I defined.

So, at A. Red collects information, doppler, speed, distance.
At B, Red sends command to event "Set:00" -> "Set your clock at 00:00"
At B, Red sends command to event "Set:2.5" -> "Set your clock at 00:02.50"
I think this is how to resolve the problem. Should have calculated it of course. But I just boost the world line.

 

[PeterDonis]

I think this is how to resolve the problem.

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?

 

[Stephanus]

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?

Ahhh, of course. Red should choose the right picture. Thanks!
[It doesn't make sense for Red to consider Blue/Green is moving. Red should convert to Blue/Green at rest]

 

[PeterDonis]

Red should choose the right picture.

"Should" for what purpose? Again, what is Red trying to accomplish?

 

[Stephanus]

"Should" for what purpose? Again, what is Red trying to accomplish?

Purpose? I don't know . I think we are too deep in math, that I forgot it's about phsyics. I'm just doing math. Okay, I'll concentrate on twins paradox.

 

[Nugatory]

Wait, how is a light year defined? Wouldn't the definition of how long a light year is depend greatly on your reference frame? Time and space both dilate; isn't the a light year's length inversely proportional to gamma of your reference frame? If you were traveling 0.9998c away from earth, wouldn't a light year to you be 52 times longer than a light year for an observer on earth?

Yes, different observers will disagree about both the distance covered and the time between departure and arrival for a given journey. Thus, you can never speak of a distance of one light-year (or any other distance, for that matter) without specifying a frame.

In this thread, we've specified in the early posts that the Earth and the endpoint of the journey are at rest relative to each other and separated by one light-year (that is, a light signal sent from Earth and reflected back from a mirror at the destination will come back two years after it is sent). We're then figuring the distances and times using a frame in which the spaceship is at rest, and comparing these with the values we have in the frame in which the Earth is at rest.

That's a reasonable enough thing to do and long as we remember that there's nothing special about any of these frames, not even the one in which the Earth is at rest. It just so happens that that frame is particularly convenient to use if you're trying to answer the question "How long will I have to wait for the ship to complete a round trip?".

 

[PeterDonis]

you can never speak of a distance of one light-year (or any other distance, for that matter) without specifying a frame.

Just to clarify what I was saying earlier, this is true, but it's also true that we can define a "light-year" as a unit of distance without specifying a frame. It's the distance light travels in one year. The frame dependence comes in when we realize that "the distance light travels in one year" refers to different sets of events in spacetime for different frames.

 

[Nugatory]

Just to clarify what I was saying earlier, this is true, but it's also true that we can define a "light-year" as a unit of distance without specifying a frame. It's the distance light travels in one year. The frame dependence comes in when we realize that "the distance light travels in one year" refers to different sets of events in spacetime for different frames.

Ah - right. Yes, that's an important clarification.

 

[Kyle.Nemeth]

Then you are considering a different scenario from the OP of this thread. In the OP, the distance was one light-year in the Earth frame.

In the original post, S_David asks how long it would take for a person on a spaceship to travel one light year, not to an observer on the Earth. This led me to believe that we want to know how long the trip takes relative to the ship observer. The original post does not explicitly state that the distance is one light year in the Earth frame, so is it implied by this portion of the question,

how long it would take for the person on the spacecraft to travel one light year,

that the distance must be 1 light year in the Earth frame since S_David is on Earth "measuring" the light year?

No, it would be to incorrectly assume that moving at c works the same as moving slower than c. It doesn't.

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.

Also, if we consider the reference frame of a light beam moving parallel to the length L₀ (L₀ is the length measured by an observer on a spaceship at rest) between two points in outer space, then in the light beams frame,

L = (√1-β²) L₀ = 0

Is this why the theory applies to light moving at c?

But SR applies just as well to things moving at c (like light) as to things moving slower than c.

Isn't light the only thing that moves at c?

This isn't a twin paradox scenario because the ship does not return to Earth (at least, not in the scenario proposed by the OP to this thread). That means that relativity of simultaneity enters into any attempt to compare elapsed time on Earth with elapsed time on the ship. In the standard twin paradox, where the ship returns to Earth, the two clocks (Earth and ship) can be compared directly at both the start and end of the trip. They can't in this scenario because the ship doesn't end up on Earth.

Is it also due to relativity of simultaneity that, for the ship's frame, the Earth and the beacon clocks are not synchronized?

At the instant the ship departs Earth, the beacon's clock reads 0.866 years (because the beacon and Earth clocks are not synchronized in the ship frame; the beacon's clock runs ahead of Earth's by this amount).

Is this because the ship is moving toward the clock at v = 0.866c, so that light from the beacon's clock reaches the ship 0.866 years earlier than light from the Earth's clock?

 

[Nugatory]

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?

The theory does not apply. The rules for travel at speeds less than the speed of light are derived starting from an assumption that is equivalent to denying the possibility of an observer moving at the speed of light (there are more mathematically rigorous ways of stating this, but that's a digression here) so they cannot apply in such a situation. If you try applying them and get a result that makes sense, that's just an accident. (If I try applying the theory that 1=0 to various math problems, every once in a while I'll find that I've lucked into the right answer, but that doesn't make it right to apply that "theory").

Isn't light the only thing that moves at c?

No. Gravitational waves also do, and it's a historical accident that we call the universe's invariant speed "the speed of light". We measured the speed of light before we discovered that the universe had to have an invariant speed and that several things including light would have to move at that speed.

Is it also due to relativity of simultaneity that, for the ship's frame, the Earth and the beacon clocks are not synchronized?

You could say that, and you wouldn't be wrong, but it's something of an uninteresting tautology. Consider what it means to say that two clocks are synchronized. You're basically saying that they both read the same at the same time - and relativity of simultaneity says that "at the same time" means different things on different frames so clocks synchronized in one frame will not be synchronized in another.

Is this because the ship is moving toward the clock at v = 0.866c, so that light from the beacon's clock reaches the ship 0.866 years earlier than light from the Earth's clock?

No. Time dilation and relativity of simultaneity are what's left after you've allowed for light travel time.

 

[PeterDonis]

In the original post, S_David asks how long it would take for a person on a spaceship to travel one light year, not to an observer on the Earth.

Yes, but the OP wasn't clear about whose frame the one light-year distance was supposed to be relative to. In subsequent discussion, it seems clear that he meant one light-year relative to Earth, not the ship. However, either case can be analyzed; the underlying principles are the same either way.

This led me to believe that we want to know how long the trip takes relative to the ship observer.

Yes; at least that was the original question in the OP. There have been others as the thread has gone on.

The original post does not explicitly state that the distance is one light year in the Earth frame, so is it implied by this portion of the question...

Yes, agreed.

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?

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.

Have you looked at any basic SR textbooks? If you haven't, you should. This is all very basic SR.

if we consider the reference frame of a light beam

There 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.)

Is this why the theory applies to light moving at c?

No. The theory applies to light moving at c because it correctly predicts the behavior of light moving at c. It just doesn't do it the way you are thinking it does. An SR textbook will explain how it actually does.

Isn't light the only thing that moves at c?

No. Anything with zero rest mass moves at c. Light is the only commonly encountered thing that does, but there are other particles in physics that also have zero rest mass (gluons and gravitons).

Is it also due to relativity of simultaneity that, for the ship's frame, the Earth and the beacon clocks are not synchronized?

Yes. But also see Nugatory's reply.

Is this because the ship is moving toward the clock at v = 0.866c, so that light from the beacon's clock reaches the ship 0.866 years earlier than light from the Earth's clock?

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. The difference in clock readings between the beacon and Earth, in the ship frame, is due to relativity of simultaneity.

 

[PeterDonis]

The theory does not apply.

Careful. The theory applies perfectly well to light and other things with zero rest mass. It just doesn't model them the way Kyle.Nemeth is thinking it does.

 

[Stephanus]

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...?

 

[PeterDonis]

If the observer is were a photon...?

An observer can't be a photon. An "observer" has to have a rest frame, and a photon doesn't.

 

[Stephanus]

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.

 

[Kyle.Nemeth]

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

, 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?

 

[PeterDonis]

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.

 

[PeterDonis]

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/

 

[Stephanus]

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.

 

[Stephanus]

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?

 

[EngWiPy]

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!

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?

 

[PeterDonis]

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?

 

[PeterDonis]

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.

 

[Stephanus]

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
.

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.

 

[PeterDonis]

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.

 

[Stephanus]

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]

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.

 

[PeterDonis]

Just the angle look different.

So what does that mean? I still don't understand what point you're trying to make.

 

[Stephanus]

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.

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.

 

[PeterDonis]

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).

 

[Stephanus]

[..]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.
V_Blue wrt Red is 0.98c
V_Green wrt Blue is 0.6c
For Red, V_Green is 0.995c. Just slighty 0.015c than V_Blue. And if you boost the diagram where Blue is at rest, than you'll see that the different/angle between V_Green and V_Blue 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]

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".