## Relativity question.

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webgrunt
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### Relativity question.

If I understand correctly objects gain mass as they approach the speed of light and the closer to the speed of light they get, the more mass they gain.

In addition to the impossible fuel requirements, it seems that any vessel approaching the speed of light would crush itself under its own gravity as it approached infinite mass.

People would have a similar problem. If it were somehow possible for some sort of antigravity to negate the effects of acceleration, a human could be weightless but that wouldn't mean massless. At double their normal mass, I'd think the heart would start to have to work harder to pump the more massive blood. Of course the human would also find it harder to control their more massive limbs as their mass increased, it would be harder for their esophageal muscles to push more massive food and water to their stomach and so forth.

My question is, is this correct? Would the mass increase be directly experienced? Or would it be more like the time differences, with the person's experience of time remaining the same while time in the relatively stationary universe speeding up from the traveller's perspective?

Pfhorrest
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### Re: Relativity question.

It's the latter, and the "mass increase" thing is actually kind of a popular misconception that's disliked by most scientists today.

It's basically an issue of mass-energy equivalence. Things that are moving faster have more kinetic energy, and so more total energy, and so more "relativistic mass". But their rest mass, which is responsible for their gravitational effects, remain the same.

But even if one were concerned with relativistic mass, from the reference frame of the object itself, it is stationary, so its relativistic mass just is its rest mass, since there is no kinetic energy involved to consider. Conversely, the rest of the universe might seem to be speeding by, and to have a much higher relativistic mass... but the fast-moving observers wouldn't thereby expect to see the rest of the universe collapse under its own gravity because of that.
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Heimhenge
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### Re: Relativity question.

Totally agree with the idea that the moving F.O.R. see's its own rest lengths and times. But I've also heard the mass increase explained as why you can't reach the speed of light. Each unit of fuel is expended against ever-increasing inertia. See for example:

http://www.exploratorium.edu/relativity/mass.html

Is this not experiencing relativistic mass increase in the moving F.O.R. ?

speising
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### Re: Relativity question.

The reason c can't be reached has different explanations for different observers. The accelerating observer will see the universe shrink until it has length 0 at c. So there's no space to get any farther.
The resting observer will see the acceleration slow down and explain this by increasing mass. (this is what the linked article explains)
Then there's the time dilation angle as well.

gmalivuk
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### Re: Relativity question.

You don't need to resort to relativistic mass to explain the inability to reach c. Momentum has a perfectly workable equation in relativity and its also infinite at c. Rockets increase the velocity of the payload by conserving momentum between it and the propellant shot out the back, and obviously as the momentum of the payload approaches infinity, the amount of momentum you'd need to shoot out the back also grows without bound.

(You can make similar momentum-only arguments for solar or laser sails, or whatever else you propose using to accelerate.)
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ijuin
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### Re: Relativity question.

However, since the mass, and therefore energy, of your propellant (and reactor fuel if propellant and fuel are not one and the same) increases by the same factor as the rest of your starship, then acceleration should remain constant within the starship’s frame of reference. Remember that the Equivalence Principle holds for ALL frames of reference—therefore the starship crew should be unable to determine their own (non-rotational) motion solely by internal measurements without observing the external universe. Thus, engines and fuel that would accelerate the starship at 10m/s^2 of shipboard time when orbiting Earth would provide equal acceleration (within the shipboard frame of reference) at 0.9999c.

This implies that, for the travelers, acceleration remains smooth all the way up to c, but the time dilation will result in the last seconds or so for the final 0.00001% will take the rest of the lifespan of the outside universe.

Eebster the Great
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### Re: Relativity question.

I think a moving object in a vacuum (whatever "moving" even means there) won't care, but an object moving relative to something else will experience a different gravitational force as a result. For instance, if you drive very fast along the surface of the Earth, your gravitational field (from the reference frame of the center of the Earth) will be distorted longitudinally and strengthened, and therefore the car will effectively weigh more.

Pfhorrest wrote:It's the latter, and the "mass increase" thing is actually kind of a popular misconception that's disliked by most scientists today.

It's basically an issue of mass-energy equivalence. Things that are moving faster have more kinetic energy, and so more total energy, and so more "relativistic mass". But their rest mass, which is responsible for their gravitational effects, remain the same.

This is not correct. Rest mass is not uniquely responsible for gravity; all elements of the stress-energy tensor come into play, including energy density (all energy, including kinetic energy), momentum density, and momentum flux.

PM 2Ring
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### Re: Relativity question.

Modern treatments of special relativity avoid the concept of relativistic mass. It can be confusing and misleading, unless you know exactly what you're doing.

For example, it can lead to erroneous conclusions like the OP's "any vessel approaching the speed of light would crush itself under its own gravity as it approached infinite mass". Similarly, an isolated body at sufficiently high speed in some frame should have enough relativistic mass to turn into a black hole. But if it's a black hole in one frame it should be a black hole in all frames. In the rest frame of a high energy cosmic ray your speed is very close to c, so why aren't you a black hole?

As gmalivuk & others have indicated, you don't need relativistic mass for relativity calculations, you can use the relativistic definitions of momentum and total energy.

There's a nice page on the Physics Stack Exchange on this topic: Why is there a controversy on whether mass increases with speed?

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### Re: Relativity question.

ijuin wrote:However, since the mass, and therefore energy, of your propellant (and reactor fuel if propellant and fuel are not one and the same) increases by the same factor as the rest of your starship, then acceleration should remain constant within the starship’s frame of reference. Remember that the Equivalence Principle holds for ALL frames of reference—therefore the starship crew should be unable to determine their own (non-rotational) motion solely by internal measurements without observing the external universe. Thus, engines and fuel that would accelerate the starship at 10m/s^2 of shipboard time when orbiting Earth would provide equal acceleration (within the shipboard frame of reference) at 0.9999c.

This implies that, for the travelers, acceleration remains smooth all the way up to c, but the time dilation will result in the last seconds or so for the final 0.00001% will take the rest of the lifespan of the outside universe.

Proper acceleration is constant, but thanks to the combination of length contraction and time dilation the travelers can subjectively go much faster than light, in the sense that a trip across the galaxy can happen in a human lifetime.

In particular, this means that there's no upper limit to the amount of time (in whatever frame) the ship can keep accelerating.
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webgrunt
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### Re: Relativity question.

Pfhorrest wrote:It's the latter, and the "mass increase" thing is actually kind of a popular misconception that's disliked by most scientists today.

It's basically an issue of mass-energy equivalence. Things that are moving faster have more kinetic energy, and so more total energy, and so more "relativistic mass". But their rest mass, which is responsible for their gravitational effects, remain the same.

But even if one were concerned with relativistic mass, from the reference frame of the object itself, it is stationary, so its relativistic mass just is its rest mass, since there is no kinetic energy involved to consider. Conversely, the rest of the universe might seem to be speeding by, and to have a much higher relativistic mass... but the fast-moving observers wouldn't thereby expect to see the rest of the universe collapse under its own gravity because of that.

Thank you, that makes it clear.

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### Re: Relativity question.

The relevant equations of motion with explanations are at http://www.math.ucr.edu/home/baez/physi ... ocket.html

It takes 3 years on a ship accelerating at 1g to go from 0.99993c to 0.9999998c.

Rapidity is the quantity that increases linearly with constant proper acceleration.
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ijuin
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### Re: Relativity question.

Doesn’t the Equivalence Principle mean that the starship and its crew should be unable to distinguish rapidity from velocity with respect to themselves?

Eebster the Great
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### Re: Relativity question.

ijuin wrote:Doesn’t the Equivalence Principle mean that the starship and its crew should be unable to distinguish rapidity from velocity with respect to themselves?

Rapidity and velocity are not the same thing. Rapidity is the hyperbolic arctangent of velocity (with speed of light = 1). From your own reference frame, your velocity is 0 by definition, and so is your rapidity, because artanh(0) = 0. But for any other speed, artanh(x) > x, so the rapidity is greater than the (normalized) velocity. At the speed of light (velocity = 1), rapidity is infinite, since the hyperbolic tangent approaches 1 asymptotically.

The point is just that if I am standing still and watch you move away from me at constant proper acceleration (in other words, it feels to you like your acceleration is constant), then in my reference frame, your rapidity will increase at a constant rate. For speeds much less than the speed of light, this is true for velocity anyway (constant acceleration means your velocity increases at a constant rate). But as you approach the speed of light, even though your proper acceleration hasn't changed, your actual acceleration (and thus the actual rate your velocity changes) from my reference frame will get smaller, as your speed asymptotically approaches the speed of light. However, your rapidity will keep increasing at the same rate it always has, up toward infinity.

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### Re: Relativity question.

ijuin wrote:Doesn’t the Equivalence Principle mean that the starship and its crew should be unable to distinguish rapidity from velocity with respect to themselves?

Why would that be the case? Every velocity (in one dimension) has an associated rapidity, w = artanh(v/c), regardless of what the velocity is relative to. You can measure your own velocity relative to something else of your choice, and then you can plug it into that equation and get the corresponding rapidity, which will be different except for v=w=0.
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ijuin
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### Re: Relativity question.

If c is 300,000,000 m/s (rounding for simplicity), and I am accelerating at a subjective 10 m/s^2, then how is it that under the Equivalence Principle I do not perceive myself reaching c after 30 million subjective seconds? Any deviation from “constant subjective acceleration equals linear subjective increase in delta-v” would constitute information about my velocity with respect to c, which by the Equivalence Principle can not be obtainable without looking outside of my starship.

Pfhorrest
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### Re: Relativity question.

You never perceive yourself reaching any nonzero velocity without reference to things outside your starship.
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p1t1o
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### Re: Relativity question.

ijuin wrote:If c is 300,000,000 m/s (rounding for simplicity), and I am accelerating at a subjective 10 m/s^2, then how is it that under the Equivalence Principle I do not perceive myself reaching c after 30 million subjective seconds? Any deviation from “constant subjective acceleration equals linear subjective increase in delta-v” would constitute information about my velocity with respect to c, which by the Equivalence Principle can not be obtainable without looking outside of my starship.

How would you percieve your speed? How would you know how fast you are going? Measure distance travelled per unit time right? Well funny things happen to the flow of time when you approach c. Now how do you tell how far you have travelled? By looking at physical "landmarks"? Funny things happen to distance when approaching c as well.

All that conspires together to mean that the answer is not all that simple.

But it helps me to consider a torchship that can accelerate at 1G indefinitely.

In such a a ship, you would approach so close to c, time dilation would get so extreme, that you could circumnavigate the known universe within a century of shiptime.

A passenger would surely note the passage of billions of galaxies over the years and calculate an extreme velocity well in excess of c.

But a stationary observer would never see the ship pass c.

**edit

But now that I think about it, what would a passenger observe at such extreme velocity? The perception of the outside universe could be severely distorted. Distances - or at least perception of distances - would be distorted so what speed *would* they calculate? Actually I have no idea.

Eebster the Great
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### Re: Relativity question.

A passenger would never calculate a velocity greater than 0 relative to himself, as Pfhorrest said. We only see other things moving relative to us. And those other things get time dilated and length contracted too. In the example of someone in a spaceship with a constant proper acceleration of 1 g, the passenger would feel a constant 1 g acceleration in his gut, and he would at first observe everything else in the universe accelerating toward them at 1 g. This is the same thing that happens when we drive a car (but hopefully you don't accelerate this fast in a car...). But as the speeds of things the passenger is looking at (say, stars in front of him in the distance) increases toward c, he will start to notice relativistic effects, just like an observer at home would notice the ship experience relativistic effects. The passenger will see the distance to these stars contract and the stars' clocks slow down. If the star had previously been rotating once a month, now maybe it's only rotating once a year, and eventually once a century or once a billion years. As this happens, the distance to the star contracts. Eventually, the distance will be so short, it will hardly take any time at all to pass it. He never observe it moving faster than c, but he will reach it sooner (in his own reference frame) than we would have naively believed possible if we didn't know how relativity worked. An observer at home will see him taking far longer to reach the same destination, but they will understand that he is aging very slowly over the course of this journey due to time dilation. Either way, the result is the same--he reaches the star before he gets senile.

By the same token, a high-energy muon has no problem flying from the upper atmosphere of the Earth all the way to the surface before it decays, even though it has a half-life of only 2.2 microseconds. That's because in the muon's reference frame, the Earth's atmosphere is only a few kilometers thick (or less), so it has plenty of time to reach the surface (or technically in the muon's frame, the Earth has plenty of time to reach it). In our reference frame, the Earth's atmosphere seems too thick for the muon to be very likely to make it all the way to the surface given its short half-life, except the muon is moving so fast that it is time dilated, so actually it should decay much more slowly than if it were at rest. Either way, there's the same result--the muon usually reaches the surface before decaying.

gmalivuk
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### Re: Relativity question.

p1t1o wrote:But now that I think about it, what would a passenger observe at such extreme velocity? The perception of the outside universe could be severely distorted. Distances - or at least perception of distances - would be distorted so what speed *would* they calculate? Actually I have no idea.

Depends how they calculate.

If it's based on individual external things, using ship-measured distances and ship-measured times, the calculated velocity would be the same v that external observers calculate.

However there's also the "mile-marker" value which I'll call "speediness" because someone already used "rapidity". It's the speed they calculate based on the proper time elapsed between external objects if known distance apart in their own frame. (In other words, it's analogous to calculating a car's speed by timing gap between mile markers on the highway.)

This value is v*gamma(v) = v/sqrt(1 - v^2/c^2).

At v = c/sqrt(2), speediness is c. That means a ship at that speed would perceive the trip from Sol to Alpha Centauri to take 4.37 years.

But at a rapidity of 10.3, which is what you get after 10 years of 1g proper acceleration, speediness is nearly 15000. Two stationary beacons a lightyear apart in their own frame would be about 35 light-minutes apart in yours and traveling at very nearly c relative to your ship, so you'd pass them in about 35 minutes.
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p1t1o
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### Re: Relativity question.

gmalivuk wrote:<snikt>

If Im understanding correctly - time and distance distortions make it so that if you try and measure the time and distance travelled, you get the same asymptote-to-c that outside observers would measure.

But if you have "landmarks" with a known distance, you calculate your velocity increasing without bound?

That seems to make sense.

***

Start Trek where the warp drive is still subject to relative time dilation:

S01E01:

Captain Kirk sets out on his 5 year voyage and never returns. Why are we so alone in the void?

FIN

Eebster the Great
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### Re: Relativity question.

Sort of. The landmarks with "known distance" will be closer together, allowing you to pass more per hour, so your "speediness" in markers/hr will increase without bound. But since the markers are no longer a mile apart, that is no longer equal to your speed in miles/hr. The "known distance" here really means "distance in a chosen inertial reference frame," since technically there is no unique "true" measure of distance for all observers.

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### Re: Relativity question.

ijuin wrote:Any deviation from “constant subjective acceleration equals linear subjective increase in delta-v” would constitute information about my velocity with respect to c, which by the Equivalence Principle can not be obtainable without looking outside of my starship.

The others have given great replies, but I'd like to mention that in SR (special relativity) velocity doesn't have the simple linear behaviour that it does in traditional Newtonian mechanics, which uses (Galilean) relativity.

In Newtonian physics, if body A is traveling in a straight line with a speed of u relative to O, our fixed reference frame, and body B is traveling (in the same line) at v relative to A, then B's speed w relative to O is given by the simple linear equation w = u + v. We can use the same equation if A accelerates, adding v to its current speed.

In SR, velocity measures the slope of a worldline in Minkowski spacetime. When the speeds are small compared to c, that means the slope is tiny, and you can get a good approximation of combined speeds by simply adding the slopes together.

Similarly, in normal Euclidean space if you have two thin wedges that both have a slope of 1/10 (1 unit vertical per 10 units horizontal) the combined slope of one wedge on top of the other is very close to 2/10 = 1/5. But to get the true value for the combined slope you need to add the angles of the wedges together. That ends up giving a combined slope of 20/99. The general formula for combining slopes u and v is (u + v) / (1 - uv). This comes from the formula for the tangent of the sum of 2 angles.

The SR formula for combining speeds in Minkowski spacetime is very similar, apart from a sign change that happens because of how time and space combine together. In natural units, where c = 1, the formula for combined speeds is w = (u + v) / (1 + uv). Eg, if u = v = 1 / 10, then w = 20 / 101. So if A is traveling at .1c relative to O, and B is traveling at .1c relative to A, then B is moving at 20c / 101 in the O frame.

Similarly, if A is traveling at .1c relative to O, and then it accelerates by .1c, boosting itself to the B frame, O will measure A's new speed to be 20c / 101.

With a little bit of calculus, we can derive a formula for the speed of A in O's frame if A is undergoing constant acceleration, that is, A experiences a constant g force. The result is v = c * tanh(aT / c), where T is the time as measured by the ship's clock, not the time on O's clock. tanh() is the hyperbolic tangent function. I won't show the derivation for that formula in this post, but you can see it here, if you're curious.

If you want to see other formulae for constant acceleration in SR, please see the classic Usenet
relativistic rocket page.