Special Relativity hw question

[mooeypoo]

Hey guys,

 

I have hw in relativity and I have no clue how to even start. I looked up all the equations, I know the PRINCIPLES (time dilation, length dilation, etc) but I just cant figure out how to ACTUALLY solve this.

 

The question:

A space traveler with 20 years to live wants to see a nebula 100 light years away. How fast must he travel?

So I know that the relative time he has (IE, the dilated time, is 20 years). That means that relative to Earth, he would travel longer than what he technically has to live.

 

I have these equations:

 

[math]x=\gamma (x'+vt')[/math]

[math]t=\gamma (t'+\frac{v}{c^2}x')[/math]

 

Where x and t relate to the 'rest' frame and x' and t' relate to the moving frame.

 

So:

 

[math]x=100 \text{light years}[/math]

[math]t'=20 \text{years}[/math]

[math]t=\gamma(20+\frac{v}{c^2}x')[/math]

 

And I need to find [math]\frac{dx'}{dt'}[/math]

 

I have another formula:

[math]U_{x}' = \frac{dx'}{dt'} = \frac{U_{x} - v}{1-\frac{v}{c^2}U_{x}}[/math]

 

Which is where I got stuck.

 

If my x = 100, then dx/dt =0 which makes my equation above -v

which is absolutely unhelpful! :\

 

Help! what am I getting wrong here.. where do I start? What am I missing? I know the principles, but I just don't get how to actually use the formulas, and I don't have any example to go on ..

 

Thanks!

 

~moo

[swansont]

By inspection it's going to be a little over 5Gamma of dilation/contraction (100 LY in 20 years at c, means the distance has to be contracted to 20 LY, but he can't travel at c)

 

The dilated distance the traveler sees is his speed multiplied by the time in his frame, which we know to be 20 years

 

[math]\frac{L}{\gamma} = vt'[/math]

 

[math]\gamma v = \frac{L}{t'} = 5c[/math]

 

from there it's algebra (I get v = .9805 c, or gamma = 5.1)

[mooeypoo]

swansont, L is the distance? what I used as "x" ?

 

If so, I'm a bit confused. My professor wrote that there's a difference between x (rest frame) and x' (moving frame). I know that in a moving frame there's the effect of length contraction, but that's an effect that is supposed to act on the objects that are moving, no? In other words, the "ship" will contract in length, not the actual distance.

 

Am I right?

 

So if that's the case, why do I have different formulas for x (rest) and x' (moving)? Does the distance *to* the nebula changing?

 

Now, other than that, I just want to make sure I understand this part: Relative to the person moving, the movement feels "normal", and we don't use relativistic calculation (hence, the calculation is simply x=vt). Only relative to an outsider at rest are we using the relativistic calculations that "correct" for the dilation of time.

Is that right?

 

Actually, that raises another point -- the person moving doesn't feel like his time is dilating, which makes sense in theory, no matter how fast he goes - he feels as if his movement is completely "galilean" mechanics.. x=vt as distance, etc. The corrections come for the relative frame at rest, where we, the people at rest, notice that it takes him longer than we thought to reach the endpoint, and, if we had a camera inside the shuttle, we'd see time dilation on his part.

Is that right?

 

(this is a bit confusing)

 

 

I'll go over the calculations after my morning routine step-by-step so I can also make sure I can re-do it when I have another similar question

 

Thanks!

 

~moo

[timo]

If so, I'm a bit confused. My professor wrote that there's a difference between x (rest frame) and x' (moving frame). I know that in a moving frame there's the effect of length contraction, but that's an effect that is supposed to act on the objects that are moving, no? In other words, the "ship" will contract in length, not the actual distance.

 

Am I right?

No. It is any space-distance and any time-distance that gets a Gamov-factor under changes of coordinate systems, not only the diameter of one object (you'd already run into problems for a scenario with 3 rockets) or the time of one clock*.

 

Does the distance *to* the nebula change?

Yes, see above.

 

Comment: Rather than transforming the distance into the frame of the traveller you can (equivalently) just transform the time into the implicitly-assumed rest-frame (implicitly-assumed because the distance given must refer to some frame). The time it takes in the rest-frame is distance/velocity. The time in the traveller's frame is the time in the rest-frame with a suitable Gamov-factor => solve for v (might have to do that numerically).

 

*: Despite having used the word: I strongly dislike this "clock runs slower" way of explaining relativity. I do not understand it. People trying to learn relativity have struggle with it (compared to how simple classical SR actually is). Crackpots use it as a starting point for spreading nonsense about "time does not go slower, only the clocks do".

[mooeypoo]

Bah, I'm confused.

 

First off, just for the sake of simplicity (and my sanity! ) please let's call the frame where the nebula is 100 light years away (our "starting" point), the rest frame, and the frame of the moving traveler the moving frame. I'm well aware those two frames can, in their own turn, be relative frames of some other frame, but for the sake of this question, my sanity, and simplicity, let's just treat 2 frames for now.. otherwise I'm just getting more confused

 

My question is this: Part of the calculation uses the "simple", calculation x=vt. That implies that there's no relation to dilation of time or contraction of space if we ONLY look at this calculation, and none other.

Swansont used this calculation for the person inside the shuttle, the moving frame (if I understood correctly), which implied to me that for the person inside the shuttle, the movement is calculated in a non-relativistic way. x=vt is non-relativistic correction.. I calculated x=vt when I was in highschool, without knowing relativity. That was my point, and my question..

 

I don't mean that relativity doesn't apply - it obviously does - I'm just trying to figure out how to know when to use which part of the equation and where. I'm trying to make sure I know how to calculate all of these in the next homework question (and I do have a few more of these) and in the exam, etc.

 

Second, if the distance changes for the moving frame, where do I use that in the calculation? In swansont's solution I don't see a representation for the x' (moving), just for L (which, if I understood correctly, represents x-rest)...

 

 

What am I missing here?

[timo]

Let's name the variables in the moving system (referring to your nomenclature) with primes. As long as you do not change coordinate systems for the relation between distance d/d' , velocity v/v' with which the distance is crossed and time t/t' it takes to cross the distance there is no difference between SR and non-relativistic cases:

(1): v = d/t

(2): v' = d'/t'

What you probably know is that the velocity v that the rocket moves in the rest frame is the same as the velocity v' of that the target approaches the rocket with in the moving frame. The time distances measured for the process and the space distances measured for the initial state for the two different frames are related to each other by

(3): [math] t = \gamma t'[/math]

and

(4): [math]d = \gamma d'[/math].

 

Given is the time t' and the distance d' so you have calculate t via (3) (=my approach) and the plug into (1) or d' via (4) and plug into the equations for the moving frame (Tom's approach).

 

I have to go now so no more details right now. Just as a comment: Tom's [math] \frac L\gamma [/math] (the whole fraction, not only nominator or denominator) is your x'; see eq. 4.

Edited April 10, 2009 by timo

[swansont]

Bah, I'm confused.

 

Welcome to relativity.

 

Here's something that helps me keep my head from exploding. Everybody has to agree on what happened, even if they don't agree on issues of simultaneity. The fixed observer (A) sees the rocket go 100 LY at some speed. The rocket passenger (B) thinks it took 20 years. But they have to agree on the physics. They reconcile this by A claiming that B's clock ran slow, and B thinking the trip was a shorter distance, both by the same factor.

 

So yes, L is the distance (the convention I'm used to for length contraction). All I've done is write down x'=v't' for the moving frame. We know the relationship between x and x' and that's useful because x is known. v and v' are the same (I was sloppy with notation here because of that, sorry), and t' is a given. The unknown terms are all related to speed.

 

We figure out that gamma = 5.1, then we know that A's clock reads 102 years, and B thinks he went 19.6 LY. (And you also discover that rounding errors have huge effects in these problems) They both agree on the speed, since that's symmetric in the problem

[mooeypoo]

Welcome to relativity.

 

Here's something that helps me keep my head from exploding. Everybody has to agree on what happened, even if they don't agree on issues of simultaneity. The fixed observer (A) sees the rocket go 100 LY at some speed. The rocket passenger (B) thinks it took 20 years. But they have to agree on the physics. They reconcile this by A claiming that B's clock ran slow, and B thinking the trip was a shorter distance, both by the same factor.

This is where I get confused,though: Isn't it enough that the clock ran slow, to explain how it took different times for (A) and for (B) ? Why do we need to explain it with the distance being shorter too?

 

Also, again, my professor explained the x' (contraction of distance) as a story about a train moving. The train itself will contract, not the distance it's moving - at least that's what I understood..

 

He also told us about an oval "whispering gallery" (only with light) that, when it moves close to the speed of light, it becomes a sphere rather than an "oval" shape.. because it contracts..

 

So.. the object ITSELF is contracting. Right? No? My brain does, right about now.

 

So yes, L is the distance (the convention I'm used to for length contraction). All I've done is write down x'=v't' for the moving frame. We know the relationship between x and x' and that's useful because x is known. v and v' are the same (I was sloppy with notation here because of that, sorry), and t' is a given. The unknown terms are all related to speed.

Okay so I think that's another aspect that confused me. The v and v' are the same? I thought that I will end up with two different velocities -- the relative velocity for (A) and relative velocity for (B), but I think that was me overcomplicating things.. it would've been 2 velocities if the traveler inside the moving shuttle was.. like.. running or something. But he's stationary compared to the moving shuttle, so I guess it's logical how both velocities are the same.

 

We figure out that gamma = 5.1, then we know that A's clock reads 102 years, and B thinks he went 19.6 LY. (And you also discover that rounding errors have huge effects in these problems) They both agree on the speed, since that's symmetric in the problem

Yeah, I don't get the idea that BOTH contraction of distance AND the slowing down of time is ahppening.. I thought only one is enough to explain this (obviously fi time moves slower, it would take the traveler longer, compared to the stationary observer..).. but .. I.. am not sure.. anymore..

 

Graah.. These equations are simple algebra, it's just the IDEA that drives me crazy. I am sure i'm not the only one, I just want to make sure I understand this now before we (probably) move on to more complicated notions within relativity.

 

And to think we didn't even touch General relativity.. oie..

 

Anyhoo, thanks! sorry for being a pain

 

~moo

[D H]

This is where I get confused,though: Isn't it enough that the clock ran slow, to explain how it took different times for (A) and for (B) ? Why do we need to explain it with the distance being shorter too?

Because both observers have to agree on the final result. Time dilation and length contraction are flip sides of the same coin.

 

Also, again, my professor explained the x' (contraction of distance) as a story about a train moving. The train itself will contract, not the distance it's moving - at least that's what I understood..

The observer on the train sees the train as having the same length at all times. Its all relative ...

 

 

Graah.. These equations are simple algebra, it's just the IDEA that drives me crazy. I am sure i'm not the only one, I just want to make sure I understand this now before we (probably) move on to more complicated notions within relativity.

The postulates are easy to understand. The math is simple algebra. The consequences are admittedly bizarre and counterintuitive. Just follow the math. Its not like they're doing anything hairy, after all.

 

And to think we didn't even touch General relativity.. oie..

They haven't touched GR yet because with GR it is the math that makes your head explode.

[mooeypoo]

Thanks a lot everyone I'm going to go on to the next question, so I hope I got the principle enough to answer it. If not, well, expect some more questions. I *think* I got it.. but.. well, it's still getting my head to spin.

 

Btw,

 

They haven't touched GR yet because with GR it is the math that makes your head explode.

That makes me feel so much better.... eeek!

 

Thanks guys

 

~moo

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Okay, I'm sorry guys, but this doesn't work, the variables just don't add up for me..

 

I represent the moving frame with ' --> x', t'

The rest frame is represented without that --> x, t

Velocities are the same for both --> v' = v

 

Here are my formulas:

 

[math]x=\gamma (x' + vt')[/math]

[math]t=\gamma (t' + \frac{v}{c^2}x')[/math]

And, from the moving reference:

[math]x'=\gamma (x' - vt')[/math]

[math]t'=\gamma (t' - \frac{v}{c^2}x')[/math]

 

And I have the speed of each frame relative to itself:

[math]x' = vt'[/math]

[math]x = vt[/math]

 

Now, my variables according to the problem:

 

[math]t' = 20 \text{ y}[/math]

[math]x = 100 \text{ Ly}[/math]

 

The x' will not be the same as x, because of length contraction.

The t' will not be the same as t, because of time dilation.

 

Swansont, I don't understand where you got your formulas from.. am I missing a formula? Am i missing a process of linking any of them together?

 

I've been playing around with this and I can't manage to do the actual process. I get it that it should be close to 5 something (because 100/20 = 5) but .. I can't do that in practice in this exercise.

 

This is where I got so far in the actual calculation:

 

[math]x=100c=vt=\gamma (x'+vt')=\gamma (x' + 20v)[/math]

[math]x'=vt'=\gamma (x-vt)=\gamma (100c - vt')[/math]

 

And since v is equal in both reference frames,

[math]v = \frac{100c}{20\gamma}-\frac{x'}{20}=\frac{x'}{t\gamma}-\frac{100}{t\gamma}[/math]

 

I, quite simply, have too many variables.

 

I may be overcomplicating myself. That is known to happen. But I don't want to just take swansont's method as-is without figuring out how and where it came from.. specially when it's such a confusing subject.

 

So.. I'm sorry for being a pain, but.. HEEELLLPPP!!!!! I'm.. like.. sobbing. Every time I think I got it, I get stuck.

 

~moo

[swansont]

This is where I get confused,though: Isn't it enough that the clock ran slow, to explain how it took different times for (A) and for (B) ? Why do we need to explain it with the distance being shorter too?

 

Because the traveler doesn't think his clock is running slow. In his frame it's fine — he's at rest, and the whole rest of the universe is moving at high speed. But they both have to be able to say that d = vt — they both have to agree that the physics works, and also that the traveler arrived at the destination just as he died.

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I, quite simply, have too many variables.

 

You've written down the general equations, and need to apply the problem conditions to them.

 

[math]x=\gamma (x' + vt')[/math]

 

That's a general transform between the two frames, i.e. how to find coordinate x in terms of x', with the frames moving at a relative speed of v, as a function of time in the transformed frame.

 

But if I don't care how it evolves in time, I can take a "snapshot" of the two coordinate systems at t=0, which tells me that [math]x=\gamma x'[/math] if we share a common origin. So you know that at the end of the trip of length L, [math]L=\gamma L'[/math], which is the general length contraction fomula

 

Or, I can see how one particular location evolves in time. Let's choose this to be the origin. Why? Because we can — choose convenient coordinate systems to make the math easier. The traveler is at the origin of his reference frame. How does that match up at the end of the trip? i.e. where is that point in the other coordinate system?

 

[math]x=\gamma (x' + vt')[/math] so x'=0, and we get [math]x=\gamma vt'[/math] , with x=100ly and t' = 20 years, which (rearranged) is the formula I used in post 2

 

[math]\gamma v = 5c[/math]

 

From there on it's algebra

Edited April 11, 2009 by swansont
Consecutive posts merged.

[mooeypoo]

K, I think I've solved it and I'll have to wait for class to ask all the rest of the questions I have on the subject and on the other questions I have in my hw, otherwise i'd be pretty much spamming this forum.

 

However, I thought you guys would get a doozy out of this one (I dont have time to solve it atm I just thought the question itself is awesome):

 

Problem 12.13 Sophie Zabar' date=' clairvoyante, cried out in pain at precisely the instant her twin brother, 500km away, hit his thumb with a hammer. A skeptical scientist observed both events (brother's accident, Sophie's cry) from an airplane traveling at [math']\frac{12}{13}c[/math] to the right [there's a figure, too, if you insist I'll put it up, ~moo]. Which event occured first, according to the scientist? How much earlier was it, in seconds?

Gotta love physics.

[timo]

It's not so awesome, actually:

- For a start, it should be clear that assuming movement of the airplane is co-linear with the distance of the twins then the orientation of airplane movement determines the order of the events while the time-distance in the new frame is independent of orientation.

- All the question is about is that you have two positions in spacetime P1=(0,0) and P2=(0,X) that get transformed to P1'=[math]\Lambda P_1[/math] = (p1t, p1x) and P2'=(p2t, p2x), respectively (1st entry is the time coordinate, 2nd is space coordinate, i.e. X=500 km, [math]\Lambda[/math] symbolizing a Lorentz-transformation). The time-distance is |p1t - p2t|, of course.

[mooeypoo]

I thought the fact I have a question with clairvoyante and a skeptic, regardless of the physics at the moment, is cool by itself.

 

Besides, remember this is a beginner's course on relativity, so obviously the problem ITSELF is going to be easy and the situation simplified... I just started laughing when I saw the clairvoyante thing and thought you'd share a laugh...

 

 

~moo