quick question on time dilation

[iNow]

I thought you made them up arbitrarily.

[D H]

I thought you made them up arbitrarily.

Correction: The original poster (theANTIcraig, not New Science) made them up arbitrarily. The specific values in the OP:

[math]v = 0.9999999999\,c[/math] (only 3 cm/sec shy of c!)

[math]L = 5\,\text{light years}[/math] (one-way distance to star, Earth frame)

 

I'm assume the spacecraft has an inertialess drive so it can change velocity nearly instantaneously without destroying the ship or crushing the pilot. In the original rest frame (i.e., the Earth frame), the ship reaches the other star in 5 years and 15.778463 milliseconds. Total time there and back is thus 10 years and 31.556926 milliseconds, plus whatever time was spent sightseeing at the other star.

 

What happens from the pilot's perspective is a different matter. Ignoring distance traversed while accelerating (very short, by assumption), that 5 light year distance to the other star shrinks to 37.1901939 light minutes. In the pilot's frame, the ship is approaching the star at 3 cm/sec shy of light speed. The outbound trip takes 37.1901939 minutes. Total trip time for the pilot: 74.3803878 minutes, plus whatever time was spent sightseeing at the other star.

 

So the faster he travels the less distance he travels?

If he goes fast enough, the 2.5 million light years to the Andromeda galaxy becomes less than the distance traversed during walk in the park.

[Motor Daddy]

Correction: The original poster (theANTIcraig, not New Science) made them up arbitrarily. The specific values in the OP:

[math]v = 0.9999999999\,c[/math] (only 3 cm/sec shy of c!)

[math]L = 5\,\text{light years}[/math] (one-way distance to star, Earth frame)

 

I'm assume the spacecraft has an inertialess drive so it can change velocity nearly instantaneously without destroying the ship or crushing the pilot. In the original rest frame (i.e., the Earth frame), the ship reaches the other star in 5 years and 15.778463 milliseconds. Total time there and back is thus 10 years and 31.556926 milliseconds, plus whatever time was spent sightseeing at the other star.

 

What happens from the pilot's perspective is a different matter. Ignoring distance traversed while accelerating (very short, by assumption), that 5 light year distance to the other star shrinks to 37.1901939 light minutes. In the pilot's frame, the ship is approaching the star at 3 cm/sec shy of light speed. The outbound trip takes 37.1901939 minutes. Total trip time for the pilot: 74.3803878 minutes, plus whatever time was spent sightseeing at the other star.

 

 

If he goes fast enough, the 2.5 million light years to the Andromeda galaxy becomes less than the distance traversed during walk in the park.

 

1. Acceleration doesn't occur instantly. E=MC^2 (Energy=Mass*The acceleration of 186,0000 mi/sec^2). C^2 is an acceleration value of 186,000 mi/sec^2. In order for an object to accelerate to ~ the speed of light (186,000 mi/sec) in, say, .5 seconds, from an initial velocity of zero, the acceleration would have to be 373,000 mi/sec^2, and at that acceleration rate, the ship would be traveling 373,000 mi/sec after 1 second. Can you tell me what the actual acceleration of the ship is, and stop talking about impossible scenarios?

 

2. How was the 5 light year distance determined?

 

3. How was the velocity of .99c determined?

 

4. Are you assuming an initial velocity of zero?

 

5. How long does it take for the ship to reach the velocity of .99c from an initial velocity of zero?

 

6. So Andromeda is 2.5 light years away. That means traveling at the speed of light (186,000 mi/sec) it would take me 2.5 years to get there. If I travel at the speed of light towards Andromeda, and the distance becomes only a walk in the park, and a walk in the park is ~1 mile, why does it take me 2.5 years to travel 1 mile at the velocity of 186,000 mi/sec?

Edited June 20, 2008 by Motor Daddy

[swansont]

1. Acceleration doesn't occur instantly. E=MC^2 (Energy=Mass*The acceleration of 186,0000 mi/sec^2). C^2 is an acceleration value of 186,000 mi/sec^2. In order for an object to accelerate to ~ the speed of light (186,000 mi/sec) in, say, .5 seconds, from an initial velocity of zero, the acceleration would have to be 373,000 mi/sec^2, and at that acceleration rate, the ship would be traveling 373,000 mi/sec after 1 second. Can you tell me what the actual acceleration of the ship is, and stop talking about impossible scenarios?

 

It's an idealized case, simplified to reduce unnecessary distractions so as to reduce confusion.

 

 

2. How was the 5 light year distance determined?

 

3. How was the velocity of .99c determined?

 

AFAIK, they were arbitrarily chosen.

 

4. Are you assuming an initial velocity of zero?

 

5. How long does it take for the ship to reach the velocity of .99c from an initial velocity of zero?

 

For purposes of the argument, we're ignoring that additional step; either you synchronize clocks at that speed or you ignore the time it takes to accelerate.

 

6. So Andromeda is 2.5 light years away. That means traveling at the speed of light (186,000 mi/sec) it would take me 2.5 years to get there. If I travel at the speed of light towards Andromeda, and the distance becomes only a walk in the park, and a walk in the park is ~1 mile, why does it take me 2.5 years to travel 1 mile at the velocity of 186,000 mi/sec?

 

As the other example showed, it wouldn't. Someone on earth would think it took you that long, but you would think the trip was much, much shorter, because your (identical) clocks wouldn't be ticking at the same rate.

 

However, to avoid more complication, let's stick with the original scenario and not have people referring to two (or more) examples.

[ydoaPs]

1. Acceleration doesn't occur instantly. E=MC^2 (Energy=Mass*The acceleration of 186,0000 mi/sec^2). C^2 is an acceleration value of 186,000 mi/sec^2.

Actually, E²=(mc²)²+(pc)². C² is NOT an acceleration. It is the speed of light (3x10⁸m/s) squared. [math]c^2={9x10^{16}}\frac{m^2}{s^2}[/math]

 

 

2. How was the 5 light year distance determined?

 

3. How was the velocity of .99c determined?

They are the measurements from a frame of reference which is at rest with respect to the Earth.

[D H]

1. Acceleration doesn't occur instantly. Can you tell me what the actual acceleration of the ship is, and stop talking about impossible scenarios?

Designate the spaceship's acceleration (as measured by an accelerometer onboard the spaceship) as [math]\kappa c[/math]. Obviously [math]\kappa[/math] must have units of 1/time. This is a hypothetical problem. I can make [math]\kappa[/math] very large. For example, if [math]\kappa = \frac{10^5}{\surd 2}\,\text{sec}^{-1}[/math] it will take a bit less than one second (coordinate time) for the ship to reach the desired speed. For [math]\kappa = \frac{10^8}{\surd 2}\,\text{sec}^{-1}[/math], the coordinate time taken to reach the desired speed is a millisecond. Now imagine a spaceship with an even larger [math]\kappa[/math]. The time elapsed and distance traversed during the acceleration phase becomes negligibly small as [math]\kappa[/math] becomes ever larger.

 

2. How was the 5 light year distance determined?

3. How was the velocity of .99c determined?

This is a hypothetical problem. The original poster made the numbers up.

• Five light years is the distance to the star as measured by observers fixed with respect to the Earth.
  
• 0.99c (actually, 0.9999999999c per the OP) is the speed of the spacecraft with respect to the Earth as measured by observers fixed with respect to the Earth. It is also the speed at which the spacecraft is moving away from the Earth (or toward the star) as measured by instrumentation onboard the spaceship.

 

4. Are you assuming an initial velocity of zero?

The spaceship starts with an initial velocity of zero with respect to the Earth.

5. How long does it take for the ship to reach the velocity of .99c from an initial velocity of zero?

This can be made negligibly small by endowing the spaceship with a very large acceleration.

 

6. So Andromeda is 2.5 light years away. ...

As swansont said, lets stick to the problem as stated by the OP.

 

So even classically velocity measurements are all relative. One of the things that comes with more advanced physics is the lack of an absolute frame of reference, and another is that velocities don't add in the way they do classically as above.

This is a very important point Klaynos raised. Einstein's theory of special relativity is not the first theory of relativity. The first such theory predates Newtonian mechanics. In "Dialogue Concerning the Two Chief World Systems", Galileo noted that that a person below decks of a ship cannot tell whether the ship is at sea or docked. The principle of Galilean relativity states that any two observers moving at constant speed and direction with respect to one another will obtain the same results for all mechanical experiments.

[Edtharan]

So even classically velocity measurements are all relative. One of the things that comes with more advanced physics is the lack of an absolute frame of reference, and another is that velocities don't add in the way they do classically as above.

You are right. At low relative velocities this discrepancy between classical mechanics and relativity is not noticeable (it does exist however). But when you get to high speeds the discrepancy is much more noticeable.

 

It is like a speedo that is 10% out. At 10km/h it will only be 1 km wrong. This is not too much of a concern as the size of the needle usually gives a greater error to the reading.

 

But at 100km/h, you will now have an error of +/-10km, far bigger than the needle and much more noticeable.

 

Now the error with Relativity is not linear like this speedo example. It is instead geometrical, so that at the relay high speeds it has so much error, our experience with the unnoticeable error at low speeds make it look like nonsense.

 

6. So Andromeda is 2.5 light years away. That means traveling at the speed of light (186,000 mi/sec) it would take me 2.5 years to get there. If I travel at the speed of light towards Andromeda, and the distance becomes only a walk in the park, and a walk in the park is ~1 mile, why does it take me 2.5 years to travel 1 mile at the velocity of 186,000 mi/sec?

First Andromeda is far more than 2.5 light years away. Andromeda is a galaxy of billions of stars. The closest star to Earth is the Sun at 8 light minutes away. The next closest is Proxima Centauri at around 4.22 light years. A whole galaxy of billions of stars can not fit between us and Proxima.

 

I think you meant 2.5 million light years.

 

5. How long does it take for the ship to reach the velocity of .99c from an initial velocity of zero?

That would depend on the rate of acceleration...

 

Oh wait this was given by DH: "I'm assume the spacecraft has an inertialess drive so it can change velocity nearly instantaneously without destroying the ship or crushing the pilot."

 

So your answer then would be: Almost Instantaneously.

 

3. How was the velocity of .99c determined?

Well if you know the rate of acceleration and know the time you accelerated for, then you can work out the speed. Or you would just compare your velocity to your starting frame of reference (Earth).

 

2. How was the 5 light year distance determined?

It was measured from Earth before the space ship set off. We can do this by several methods, but for a star so close, the best way is probably by Parallax

 

That 5 light year distance is only valid for the reference frame that the measurement was made in. If you change your reference frame then you have to take the measurement again, or using Relativity calculate it from the original measurement and your new reference frame.

 

1. Acceleration doesn't occur instantly.

Yes, and he said: "Almost Instantly". So that was a Strawman.

 

Can you tell me what the actual acceleration of the ship is, and stop talking about impossible scenarios?

It is currently imposible for us to build a space ship that could take us to another star and return. So does this mean that we should stop talking about space ships too?

 

Stop being pedantic for the sake of being pedantic. These are Thought Experiments. They are theoretical situations designed to demonstrate the principles being discussed. We are not Rocket Scientists (well some of us might be ). In Fact, most Rocket Scientists would not have to use much Relativity in their day to day work. We are talking Physics, not the engineering requirements to send rockets to a distant star (or even just to get them into orbit).

 

In order for an object to accelerate to ~ the speed of light (186,000 mi/sec) in, say, .5 seconds, from an initial velocity of zero, the acceleration would have to be 373,000 mi/sec^2

Hang on!

 

You just computed the acceleration and then just asked us to compute it. So are you just asking us to check you maths?

[D H]

In Fact, most Rocket Scientists would not have to use much Relativity in their day to day work. We are talking Physics, not the engineering requirements to send rockets to a distant star (or even just to get them into orbit).

Bingo. I need to augment my meagre Science Forums salary (exactly zero dollars and zero cents per post) with something a bit more substantial. So I moonlight from Science Forums as a rocket scientist. Relativity does not come up very often for the simple reason that the errors resulting from assuming Newtonian mechanics are very tiny compared to the errors induced by uncertainties in atmospheric drag, spherical harmonics coefficients for planetary gravity models, tidal effects, solar radiation pressure, and spacecraft thrust.

[Motor Daddy]

Just hypothetically, my car accelerates to 186,000 mi/sec in one second, and travels 93,000 miles during that one second. Oh, did I mention the engine only has 400 lb-ft of torque at 3,000 RPM??

 

No need for actual numbers, we are playing pretend physics with pretend vehicles, that accelerate against the laws of physics.

 

But that shouldn't be a problem for most of you, UNLESS, someone else says it, then you pick it apart! :rolleyes:

[Cap'n Refsmmat]

These aren't against the laws of physics. It's just a really awesome spaceship. It's not breaking any physical rules like your car traveling the speed of light.

[swansont]

I split off the train example to a separate thread: http://www.scienceforums.net/forum/showthread.php?t=33673

 

Let's make the numbers pertain to the example we already have.

[Motor Daddy]

These aren't against the laws of physics. It's just a really awesome spaceship. It's not breaking any physical rules like your car traveling the speed of light.

 

So accelerating to .99c "almost instantly" is not breaking the laws of physics?

 

What is the acceleration rate of that scenario? If that acceleration rate is maintained for 1 second, what will the velocity be? How about two seconds? three? 10 minutes?

 

Surely you can maintain that acceleration rate (same force) for those time periods? If not, why not?

[swansont]

So accelerating to .99c "almost instantly" is not breaking the laws of physics?

 

Nope. Difficult, to be sure, but it violates no laws.

[Cap'n Refsmmat]

http://www.scienceforums.net/forum/showpost.php?p=417183&postcount=31

 

You could try to maintain the acceleration rate for more than a few milliseconds, but you'd find yourself only asymptotically reaching c. Your acceleration would slow down and you'd be going insanely fast, but the energy required to reach c is essentially infinite. You could never maintain that acceleration for more than a few milliseconds (and a few milliseconds is all we need to reach our desired speed).

[freezy]

What happens from the pilot's perspective is a different matter. Ignoring distance traversed while accelerating (very short, by assumption), that 5 light year distance to the other star shrinks to 37.1901939 light minutes. In the pilot's frame, the ship is approaching the star at 3 cm/sec shy of light speed. The outbound trip takes 37.1901939 minutes. Total trip time for the pilot: 74.3803878 minutes, plus whatever time was spent sightseeing at the other star.

 

It seems that we have come to very different results from our separate calculations. I'm quite sure that my calculations are the ones in error. If you would be so kind, I'd appreciate corrections. (or at least if you could show your work so I can see where I went wrong)

 

I'd appreciate it!

[D H]

It seems that we have come to very different results from our separate calculations. I'm quite sure that my calculations are the ones in error. If you would be so kind, I'd appreciate corrections. (or at least if you could show your work so I can see where I went wrong)

 

I'd appreciate it!

I used a velocity of 0.9999999999 c, as posited in the original post. Somebody keeps changing that to 0.99 c, and this change has become a virus. The gamma with [math]v=0.9999999999\,c[/math] is about [math]10^5/\surd 2[/math]. With [math]v=0.99\,c[/math], the gamma is only 7.09 or so.

 

5 years (plus a tiny fraction) times that gamma (10⁵/sqrt(2)) is 37.19 minutes.

 

Here is your error:

From the pilot's rest frame:

[math]t'=\sqrt{1-v^2/c^2}[/math]

[math]t'=\sqrt{1-(.9999999999c^2/300000}[/math]

[math]t'=0.0000141421...[/math]

The first equation has units of time on the left-hand side but the right-hand side is unitless. Something is wrong here. You need something with dimensions of time on the right-hand side. This "something" is the time taken from the perspective of an observer fixed with respect to the Earth. Dimensional analysis is a very powerful technique for checking for errors such as these ... and for debunking crackpots!.

 

Off-topic but very important concept:

I use dimensional analysis all the time as a sanity check on my own work. Getting units wrong is one of the most common mistakes newbies make.

 

Back on topic:

There is a problem in the second equation as well. Since [math]v=0.9999999999\,c[/math], [math]v/c=0.9999999999[/math]. Thus

[math]t'=t\sqrt{1-0.9999999999^2}[/math]

 

It is easier to express 0.9999999999 as 1-10^-10. Then

[math]\sqrt{1-\left(\frac v c\right)^2} = \sqrt{1-(1-10^{-10}(2-10^{-10}))} = \sqrt{2\cdot10^{-10}(1-0.5\cdot10^{-10})} \approx \surd 2\cdot10^{-5}(1-0.25\cdot10^{-10})[/math]

This expression multiplied by 5 years (and 15.778463 milliseconds) leads to the 37.19 minute trip as measured by the pilot.

Edited June 21, 2008 by D H

[Motor Daddy]

Let me get this straight.

 

The pilot traveled a distance of 5 light years in 37.19 minutes?

 

It takes light 5 years to travel the distance of 5 light years, but it only takes the pilot 37.19 minutes?

 

How many times does his heart beat during the trip if his pulse is normally 60 beats per minute?

Edited June 21, 2008 by Motor Daddy

[ydoaPs]

Let me get this straight.

 

The pilot traveled a distance of 5 light years in 37.19 minutes?

 

It takes light 5 years to travel the distance of 5 light years, but it only takes the pilot 37.19 minutes?

 

How many times does his heart beat during the trip if his pulse is normally 60 beats per minute?

 

The distance, for the traveler, is NOT 5ly; It is significantly less.

[Motor Daddy]

The distance, for the traveler, is NOT 5ly; It is significantly less.

 

How many times does his heart beat during the trip?

 

Why is the distance shorter for the traveler? The distance is 5 light years. Light takes 5 years to travel the distance.

 

I can confirm the distance is 5 light years, as I laid down a measuring tape the entire distance. The pilot is traveling along the tape, so there is no confusion on the distance, it is 5 light years. It is already measured.

[Cap'n Refsmmat]

How many times does his heart beat during the trip?

As many times as you'd expect for the half-hour trip calculated by D H.

 

Why is the distance shorter for the traveler? The distance is 5 light years. Light takes 5 years to travel the distance.

Because at high speeds, the person traveling at the high speed experiences length contraction. More info available here and here.

 

I can confirm the distance is 5 light years, as I laid down a measuring tape the entire distance. The pilot is traveling along the tape, so there is no confusion on the distance, it is 5 light years. It is already measured.

As the pilot flies along the measuring tape, he utilizes his ultra-high-speed camera to take a picture of your measuring tape and compare it to his calibrated meterstick on board the spaceship. His meterstick will reveal that the tape has somehow been shrunk -- ten meters suddenly become less than one.

 

You, on the other hand, will think the tape is perfectly normal and his meterstick has shrunk (not grown). Who's right? Both of you.

 

It is counterintuitive, yes. It makes no sense, yes. But it's the way the universe works.

[D H]

How many times does his heart beat during the trip?

The pilots heart beats about 2600 times.

 

Why is the distance shorter for the traveler?

Because the speed of light is the same in all inertial frames.

The distance is 5 light years.

The distance is 5 light years in the Earth frame. The distance is a lot less (only 37.19 light minutes) to the pilot of the spaceship. You may not like it, but this is a fact. Distance and time are relative. While your Aristotelean view of the universe may be comforting and familiar, it is also very archaic and downright wrong when erroneously extrapolated applied to very high velocities.

[Edtharan]

Let me get this straight.

 

The pilot traveled a distance of 5 light years in 37.19 minutes?

You are mixing frames of reference here.

 

The Pilot (one frame of reference) travels 5 light years (from the frame of reference of Earth). This is what is wrong.

 

The people on Earth see the Pilot travel 5 light years in just over 5 year.

 

However, form the Pilots from of reference he does not travel 5 light years and it takes significantly less time than 5 years.

 

Why is the distance shorter for the traveler? The distance is 5 light years. Light takes 5 years to travel the distance.

Because the Pilot is in a different frame of reference to the people of Earth.

 

What we have been (very patiently) trying to explain is that when you have accelerated you are in a different frame of reference to someone who is in the initial frame of reference and hasn't accelerated.

 

How many times does his heart beat during the trip?

Well assuming that the pilot is quite fit and has a hear beat around 60 beats per minute this would mean his heart beats around 2231.4 times during the trip.

 

However, the people left of Earth, their heart would beat around 30,693,600 times (for around 5 years).

 

This means that the Time for the pilot has slowed down. The Pilot sees everybody else as moving faster. And everybody else sees the Pilot moving slowly (inside the ship - not that the ship is moving slowly through space, just that time for the ship has slowed down).

 

So accelerating to .99c "almost instantly" is not breaking the laws of physics?

no. You are not exceeding the speed of light and your acceleration is not infinite.

 

We are talking Physics, not the engineering requirements to send rockets to a distant star (or even just to get them into orbit).

 

Whether you could design an engine to withstand the pressures to generate this kind of acceleration is an engineering problem, not a physics problem (although you would use physics to solve the engineering problem).

 

What is the acceleration rate of that scenario? If that acceleration rate is maintained for 1 second, what will the velocity be? How about two seconds? three? 10 minutes?

We did not specify a particular rate of acceleration. We specified an initial speed (0c) and a target speed (0.9999999999c) and a period of time (almost instantaneously). We did not say that once we reached that speed we kept on accelerating.

 

Surely you can maintain that acceleration rate (same force) for those time periods? If not, why not?

Actually to achieve a constant rate of acceleration over the specified period of time, we would need to have an increasing amount of thrust. The faster you go the more mass (not rest mass which is a slightly different thing) according to the E=MC^2 equation (moved around to be M = E / C^2). The more mass you have the more thrust is needed to accelerate you at the same rate according to A = F/M (Acceleration = Force / Mass ).

 

So the faster the ship goes the more massive it becomes, but the more massive it becomes the more force is needed to achieve the same rate of acceleration.

 

If you work these out for accelerating a ship up to the speed of light, it turns out you need an infinite amount of energy to accelerate something up to the speed of light (interestingly this only applies to accelerating something to the speed of light, if you create something moving at the speed of light - like a photon - this is different and does not take an infinite amount of energy, but the object must have 0 rest mass, and that if you have 0 rest mass you can only be created travelling at the speed of light ).

 

So, back to the example...

 

Even though our "hypothetical" ship accelerated to near light speed almost instantaneously, this rate of acceleration was either non uniform (that is our rate of acceleration reduced as we got faster) or we increased the amount of thrust to compensate for the increase in mass.

 

But, if we were to try to continue to increase our thrust and keep accelerating, then the amount of mass increase continues exponentially to the point where we would need an infinite amount of thrust to go faster. As infinities are non physical (they break the laws of physics) we can't use them and we can not accelerate to the speed of light.

 

Oh, and if you think that we can't know this because we have never accelerated anything up to that kind of speed, well...

 

We have. In particle accelerators, these kinds of velocities are achieved every day. They use magnetic and electric forces to push and pull the particles like electrons up to these phenomenal speeds.

 

We know that the particle is gaining mass, as we can detect that in the amount of energy needed to increase it's speed further (as I was saying the faster you go the more mass, the more mass the more energy needed to accelerate at the same rate) and also in that the accelerators are curved (usually) and we can detect the amount of force needed to make the particle turn around the corners.

 

So through these direct experiments, we absolutely know that when you accelerate an object it gains mass and as it gains mass it requires more energy to keep the same rate of acceleration.

 

We can measure these and draw a graph. When you plot that graph we can see that to achieve a velocity of light speed you would need an infinite amount of energy to do so.