# Another speed of light question

1. Aug 7, 2008

### orgmark

Say I'm in a spaceship travelling 0.999999999999999 c
I'm a bit curious as to how I got going that fast. I at this
point decide someone should be steering the ship. I
walk form the back of the ship to the front. What
happens when I attempt to do this?

2. Aug 7, 2008

### Integral

Staff Emeritus
You arrive at the back of the ship, turn around and walk to the front. Meanwhile you note that since the food is so terrible, not only is your mass not nearing infinite, you have actually lost weight.

3. Aug 7, 2008

### phja

basically, everything would apear normal from your point of view (referance frame). you could walk around just like you can on earth. in fact the only way you can deduce you are moving at 99.99999% speed of light is to compare it to another referance frame but even then you don't know if your moving at this speed, or the comparitive referance frame is.

so, absolute motion cannot be detected without something to compare it to. eg- the only way to know your aeroplane is moving is to look down at the ground.

4. Aug 7, 2008

### Fredrik

Staff Emeritus
If you are concerned that your walking speed v should be added to the ship's speed u to get the total speed u+v, which may be >c, then the answer is that your speed in the frame where the ship is moving with speed v isn't u+v, it's (u+v)/(1+uv/c2) and that's always <c unless u=v=c.

5. Aug 7, 2008

### phja

exactly, and from your perspective, your only moving at walking speed, the ship isn't even moving (well, you can't determine if it's moving without an outside referance frame, so you are completly valid in saying the ships not moving).

6. Aug 7, 2008

### DaveC426913

The key points from all the above arguments are:
1] within that reference frame, nothing is out the ordinary, time passes as usual; there is no way to claim that the observer is doing any speed; he might as well be stopped and the planets are rushing past him
2] relativistic velocities do not simply add. In a very sloppy nutshell, 0.99999c+0.00001c does NOT equal 1.0c, it equals 0.999990000...01c.

7. Aug 7, 2008

### orgmark

Doesn't an object's mass increase as it approaches the speed of
light? Would I be able walk to the front of the ship if my mass
was infinite? I have read enormous amounts of energy are required
to accelerate an object moving at these velocities.

8. Aug 7, 2008

### DaveC426913

Again, the observer will experience nothing untoward. It is only when compared to an external frame of reference that the changes will be apparent.

9. Aug 8, 2008

### MeJennifer

10. Aug 8, 2008

### Staff: Mentor

Your speed with respect to yourself is zero and is always zero. So you never notice any mass increase.

11. Aug 8, 2008

### MeJennifer

Indeed the proper mass, an Lorentz invarant quantity by the way, does not depend on relative speed. Relativistic mass however does but this quantity is not Lorentz invariant.

12. Aug 9, 2008

### Nickelodeon

I would have thought that you would find it extremely difficult to walk to the front due to huge energy requirements to do so. If you happened to be at the front and walked to the back then you would find it easy.

13. Aug 9, 2008

### Staff: Mentor

It doesn't take a greater-than-normal amount of energy in your own frame.

14. Aug 9, 2008

### DaveC426913

You and I are on a spaceship called Earth, which is moving at .99c relative to something. This is identical to the siutation in the smaller spaceship. We don't find it harder to move West than East.

15. Aug 10, 2008

### Nickelodeon

Agreed, but we do find it more difficult to go up rather than down and that is the "something" that we are relative with in our reference frame

16. Aug 10, 2008

### Nickelodeon

Could there be any chance that reference frames are specific to the region of space that you are in at the time? In this scenario, you are pushing your own frame through the natural regional specific frame which resists anything trying to go faster than 'c'

17. Aug 10, 2008

### Staff: Mentor

A reference frame is nothing more than a coordinate system, it is a useful mathematical tool, not a physical entity. There is no such thing as a "natural frame", and mathematical tools don't "resist" things.

The whole point of relativity is that the laws of physics are the same in any reference frame. In other words, you can choose whatever coordinate system you like, you will obtain all of the same experimental outcomes.

18. Aug 10, 2008

### matheinste

Hello Nickllodeon.

Quote:-

----Agreed, but we do find it more difficult to go up rather than down and that is the "something" that we are relative with in our reference frame -----

Don't forget up and down are themselves relative. In the case of the earth, an upward pointing vector relative to the earth's surface remains a downward pointing vector relative to the earth but relative to te sun it will chang direction and points in another direction as the eargth moves relative to the sun. Up and down are relative to the earth's surface in our normal use of the words. Roughly speaking, take a vector pointing at the sun at mid day, we call it upward pointing. Take the same vector at midnight, still pointing towards the now invisible sun, we would then call this same vector downwards, pointing in the earth's frame. Of course a vector pointing directly towards the earth's centre, downwards, remains so at all times. We find moving upwards along this upward pointing vector, with respect to the earth, more difficult because gravity is "acting" in our downward direction all the time. The wording may not be quite rigorous but i am sure will get the idea.

In the case of the spacecraft, movement within it forward and back, up or down has no relevance if there is no force acting upon it, which will be the case in an inertial frame.

Matheinste.

19. Aug 10, 2008

### Nickelodeon

So what is stopping orgmark going faster than light as he walks to the front of the spaceship?

20. Aug 10, 2008

### Varnick

The relativistic formula for velocity addition is different to Newtonian Mechanics, preventing a speed ever being equal to C.

$$v_1 + v_2 \neq v_3$$