Hypothetically If you were in a spaceship going 0.99999999999 c and you tried

[battlestar08]

Hypothetically If you were in a spaceship going 0.99999999999...c and you tried...

Hypothetically, If you were in a spaceship going 0.9999999999999...c and you tried to walk forward up to the the end of the craft, what would happen.

 

[Vanadium 50]

You'd walk forward to the end of the craft.

 

[Fredrik]

Velocities add according to

u\oplus v=\frac{u+v}{1+uv}

If the velocites aren't parallel, the formula is

\vec u\oplus\vec v=\frac{1}{1+\vec u\cdot\vec v}\bigg(\vec u+\vec v+\frac{\gamma_{\vec u}}{1+\gamma_{\vec u}}\vec u\times(\vec u\times\vec v)\bigg)

This is in units such that c=1.

 

[DaleSwanson]

As long as you aren't accelerating you won't ever notice any difference from speed in your own frame of reference. From the point of view of something that is traveling at 0.99c now relative to Earth, Earth is traveling at 0.99c. So right now, you are traveling 0.99c.

 

[Phrak]

Hypothetically, If you were in a spaceship going 0.9999999999999...c and you tried to walk forward up to the the end of the craft, what would happen.

If you didn't stop, you would smash into the forward bulkhead at the tremendous speed of ...99999999999c, and then break for a pleasant lunch of space snacks before attempting the rear bulkhead.

 

[lightarrow]

Hypothetically, If you were in a spaceship going 0.9999999999999...c and you tried to walk forward up to the the end of the craft, what would happen.

That, with respect to those who measure as 0.9999999999999...c your spaceship speed, you are moving at 0.99999999999999...c
(Note that I have added a "9", just to mean that your speed is slightly greater than the spaceship's speed).
Your speed will never reach (or overtake) c.

 

[Fredrik]

We can be more exact than that.

u=0.9999999999999
v=6*1000/3600/299792458=5.559401586635867492926278574582e-9

(That v is 6 km/s).

(u+v)/(1+u*v)=0.99999999999990000000111188031115

 

[Naty1]

Post three is from the frame of an external observer. You could also move at .99c from rear to front in the spaceship frame...and all would appear normal inside the sapce ship...

 

[jambaugh]

In Galilean relativity (Newtonian physics) we change velocity frames by adding velocities which is an affine transformation.

In SR we change velocity frames via (hyperbolic) pseudo-rotations which are linear transformations. To compose velocity frames you don't add velocities by rather the boost parameter (B) which parametrizes the pseudo-rotation.

x' = x cosh(B) + t sinh(B)
t' = x sinh(B) + t cosh(B)

Each boost you add to the previous gives you just the same situation as before, you seem to be standing still. It just changes you relative to someone else and always the relative velocity between two objects is less than c.

Here is a good analogy using the case of true rotations.

Given the radius of the Earth two people cannot be farther (in terms of straight-line distance) than 2R from each other. Suppose you are standing 1.999999999999999999999999 R away from home and you take a step forward... what happens?

What happens is that you took one step forward just as if you were at home. You need to do the math to see how that changes your distance from home but it doesn't make you pass the maximum distance because you aren't walking in a straight line but rather along a circle around the Earth. In short you are rotating rather than translating.

Similarly you don't boost velocities "along a straight line" but rather "pseudo-rotate" along a hyperbola. The hyperbola only looks like a hyperbola when we draw it on Euclidean paper (rather than in pseudo-Euclidean Minkowski space). Just as the surface of the Earth (minus terrain variations and weather) looks the same everywhere on its surface the hyperbolic surface that space-time 4-velocities stay on looks the same everywhere. Your velocity is not a property of you but is rather a relationship between you and another object. You can pick an arbitrary "home" frame but anything you see or predict should not depend on the choice of that home frame.

 

[HallsofIvy]

Hypothetically, If you were in a spaceship going 0.9999999999999...c and you tried to walk forward up to the the end of the craft, what would happen.

If you mean that you are in a spaceship which is going at 0.999999999...c relative to some observer (I think Phrap was (perhaps facetiously) interpreting this as you moving at 0.99999999...c relative to the spaceship) then you are moving very slowly relative to the spaceship and so nothing special will happen- relative to itself, the spaceship is stationary.

You can test that by getting up and walking across the room: there surely exist some frame of reference relative to which you are moving at 0.99999999...c right now!

 

[matheinste]

Hypothetically, If you were in a spaceship going 0.9999999999999...c and you tried to walk forward up to the the end of the craft, what would happen.

Vanadium 50 answers this in #2.

One of the fundamental points of relativity, Newtonian, Einsteinian etc. is that you cannot distinguish between systems in relative inertial motion. So in the original question you could enter any value for the relative velocity or you may as well say motionless. If you are on Earth in a room and ask what would happen if I tried to walk towards the door what would the answer be? To paraphrase Vanadium's reply, you would walk towards the door.

Matheinste.

 

[DaveC426913]

Now, if Earth observed you on the other hand ... they'd see a spaceship flashing past at .999999999c and inside, they'd see you virtually frozen in time. It would take you so long to walk to the front of the ship that they'd measure your speed as a mere .9999999999c. This is how the two frames are reconcilable.