Another speed of light question

[orgmark]

Say I'm in a spaceship traveling 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?

 

[Integral]

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.

 

[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.

 

[Fredrik]

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.

 

[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 completely valid in saying the ships not moving).

 

[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.

 

[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.

 

[DaveC426913]

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.

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

 

[MeJennifer]

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.

 

[russ_watters]

Doesn't an object's mass increase as it approaches the speed of
light?

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

 

[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.

 

[Nickelodeon]

I
walk form the back of the ship to the front. What
happens when I attempt to do this?

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.

 

[Dale]

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

 

[DaveC426913]

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.

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.

 

[Nickelodeon]

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.

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

 

[Nickelodeon]

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

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'

 

[Dale]

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'

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.

 

[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.

 

[Nickelodeon]

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.

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

 

[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

 

[ZapperZ]

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

May I suggest you and the OP check the hyperphysics webpage first and see if you can understand the description given there.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html#c2

Note how, for v<<c (i.e. our "normal" situation), we get back our beloved rule for velocity addition. This shows that our regular velocity addition is only a "special case" for a more generalized description on how we add velocity.

Zz.

 

[Dale]

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

ZapperZ is correct. It is not that it would take the guy too much energy to walk to the front of the spaceship, it is simply a result of the Minkowski geometry of spacetime. In Minkowski geometry velocities add such that the velocity is always less than c in any frame, regardless of how fast the ship is going in that frame or how fast he is walking in the ship's frame.

 

[Nickelodeon]

May I suggest you and the OP check the hyperphysics webpage first and see if you can understand the description given there.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html#c2

Note how, for v<<c (i.e. our "normal" situation), we get back our beloved rule for velocity addition. This shows that our regular velocity addition is only a "special case" for a more generalized description on how we add velocity.

Zz.

The examples given talk about observed phenomena at high speeds. Velocity measurement is governed by the time information takes to reach us so no way will any object be observed going faster than light in any direction it happens to be moving.
I think the problem comes when switching from observed speeds to real speeds. Are you sure the two need necessarily be the same?

 

[Dale]

I think the problem comes when switching from observed speeds to real speeds. Are you sure the two need necessarily be the same?

You misunderstand relativity. All of the relativistic effects remain even after accounting for the finite travel time of light.

Relativity is not about optical illusions, and all observers are considered intelligent enough to realize that the time that they receive a distant light signal is not the same as the time that the signal was emitted, and correctly account for the delay.

 

[Nickelodeon]

You misunderstand relativity. All of the relativistic effects remain even after accounting for the finite travel time of light.

Relativity is not about optical illusions, and all observers are considered intelligent enough to realize that the time that they receive a distant light signal is not the same as the time that the signal was emitted, and correctly account for the delay.

So if relativistic effects remain, then the pilot's mass will be approaching infinity in reality. Presumably, the energy required to accelerate him and get him to the front of the spaceship will also be approaching infinity. This is not what you infered (or I interpreted) in one of your previous threads. eg.

ZapperZ is correct. It is not that it would take the guy too much energy to walk to the front of the spaceship, it is simply a result of the Minkowski geometry of spacetime.

 

[Dale]

This is one reason why I don't like the concept of "relativistic mass". I never use that concept myself and I do not recommend its use.

When most modern physicists speak of mass they mean the Minkowski norm of the http://en.wikipedia.org/wiki/Four-momentum" , which is a Lorentz invariant. The pilot's mass is most definitely not increasing in the pilot's frame and therefore, for the pilot it does not take any more effort to walk to the front than normal.

 

[lightarrow]

So if relativistic effects remain, then the pilot's mass will be approaching infinity in reality. Presumably, the energy required to accelerate him and get him to the front of the spaceship will also be approaching infinity.

I'm in the ref. frame of a beam of neutrinos close to the Earth; I'm seeing you, in this exactly moment, traveling past me at 0.999999999c. Let me have a precise look: ...hmm...no, no, your mass doesn't seem very different from usual (unless you ate too many ice creams in the last hours...).
What they wrote you is the reality, not theories. Re-read them and if it's not enough, do it again.

 

[DaveC426913]

So if relativistic effects remain, then the pilot's mass will be approaching infinity in reality.

Stop right there.

There is no such thing is an objective frame of reference. There is no "reality versus "less than reality" in the sense you mean it.

The whole point of relativity - the very core - is exactly this: no frame of reference is preferred over any other. i.e. the pilot's frame of reference is equally valid. As far as he's concerned, he is stationary and the rest of the universe is moving. If he looked out his window, he'd see the Earth whizzing past him at .99c. If he measured the mass of a guy walking around on Earth, he'd measure the guy's mass as very high and would wonder how the guy can walk around at all.

Neither of them is wrong. Neither of the is seeing an illusion. Neither of their experiences are any less real.

 

[DaveC426913]

Let me have a precise look: ...hmm...no, no, your mass doesn't seem very different from usual

Actually, it would.

 

[lightarrow]

Actually, it would.

Are you talking about invariant mass?

 

[DaveC426913]

Are you talking about invariant mass?

No. Unless I'm mistaken though, an external observer looking at the craft as it whizzes by would be measuring its relativistic mass.

 

[cosmik debris]

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.

I suppose the food is overcooked from the Unruh radiation?

 

[Nickelodeon]

Stop right there.

There is no such thing is an objective frame of reference. There is no "reality versus "less than reality" in the sense you mean it.

The whole point of relativity - the very core - is exactly this: no frame of reference is preferred over any other. i.e. the pilot's frame of reference is equally valid. As far as he's concerned, he is stationary and the rest of the universe is moving. If he looked out his window, he'd see the Earth whizzing past him at .99c. If he measured the mass of a guy walking around on Earth, he'd measure the guy's mass as very high and would wonder how the guy can walk around at all.

Neither of them is wrong. Neither of the is seeing an illusion. Neither of their experiences are any less real.

Doesn't this suggest that faster than light travel, relative to something, lightarrow's beam of neutrinos, for instance (see prev thread). is possible. This would contravene 'c' being tops.

 

[lightarrow]

Doesn't this suggest that faster than light travel, relative to something, lightarrow's beam of neutrinos, for instance (see prev thread). is possible. This would contravene 'c' being tops.

For which reason should it suggest ftl travels? It doesn't suggest it at all. They've given you the formula:

V = (v1 + v2)/(1 + v1*v2/c^2)

which always gives a value < c, even if you re-iterate it as many times as you like.

 

[Dale]

No. Unless I'm mistaken though, an external observer looking at the craft as it whizzes by would be measuring its relativistic mass.

I think you are mistaken. How would you measure the mass of a moving object? I think the only thing you could measure is the momentum and then you would have to calculate the mass. At that point you can choose if you want to calculate the rest mass or the relativistic mass.

The only ways I can think of to measure mass require an apparatus at rest relative to the mass and therefore measure rest mass.

 

[lightarrow]

I think you are mistaken. How would you measure the mass of a moving object? I think the only thing you could measure is the momentum and then you would have to calculate the mass. At that point you can choose if you want to calculate the rest mass or the relativistic mass.

The only ways I can think of to measure mass require an apparatus at rest relative to the mass and therefore measure rest mass.

You can measure momentum, energy, kinetic energy, what you want. The point is that what they call "relativistic mass" has already a name: "energy".

MASS has NOTHING to do with it.

 

[Nickelodeon]

For which reason should it suggest ftl travels? It doesn't suggest it at all. They've given you the formula:

V = (v1 + v2)/(1 + v1*v2/c^2)

which always gives a value < c, even if you re-iterate it as many times as you like.

This formula is fine(ish) for calculating relative observed speeds but not much use for un-observable speeds.
This following text comes from another post on this forum which I thought illustrates the point I am trying to make.

Don't confuse what can be called the "3rd party separation rate" with relative speed.

Imagine this situation: Spaceship A is 1 light year to the west of me and travels at a speed of 0.9999c towards me; Spaceship B is 1 light year to the east of me and travels at a speed of 0.9999c towards me. What's their distance apart according to me? 2 light-years. When do they reach me? In about 1 year. So the separation distance decreases at a rate of about 2c according to me.

Of course the relative speed of spaceship B as measured by spaceship A is still less than c.

Now if we go for these two spaceships going passed the Earth and continuing on their way they will not be able to see each other although we still will.

Going back to the original post on this thread. Noone on Earth will see the pilot walking towards the front of the spaceship assuming his 'rate of separation' from us exceeds c. We will see him again when his sits down at the controls.

 

[matheinste]

Hello Nickleodeon.

Quote:-

---Now if we go for these two spaceships going passed the Earth and continuing on their way they will not be able to see each other although we still will. ---

This is incorrect.

Quote:-

---Noone on Earth will see the pilot walking towards the front of the spaceship assuming his 'rate of separation' from us exceeds c. We will see him again when his sits down at the controls. ---

This is also incorrect.

The speed of an observer relative to another observer never exceeds c in SR. However their speed of separation, that is the rate of increase of the distance between them, as seen by a third observer, may be up to nearly 2c. But to these two, whose separation distance is increasing at up to 2c, each will measure the other as going at less than c.

Matheinste.

 

[DaveC426913]

If we were to look in the porthole of the spaceship as it passed Earth, we would see the pilot moving so slowly that he would appear to be virtually frozen in time - i.e. walking toward the front of the ship extreeeeemely slowly. If we were to measure the pilot's speed relative to us, it would not exceed c for that reason.

 

[MeJennifer]

If we were to look in the porthole of the spaceship as it passed Earth, we would see the pilot moving so slowly that he would appear to be virtually frozen in time

What we actually see and measure rather than calculate using reference frames is that the pilot is going faster than us when the spaceship approaches us and going slower than us when the spaceship passes us. One can use the relativistic Doppler formula to calculate the exact ratios.

 

[Nickelodeon]

The speed of an observer relative to another observer never exceeds c in SR. However their speed of separation, that is the rate of increase of the distance between them, as seen by a third observer, may be up to nearly 2c. But to these two, whose separation distance is increasing at up to 2c, each will measure the other as going at less than c.

Matheinste.

I don't disagree with this except for the fact that after the speed of separation is greater than c the wavelength becomes too great to do any measuring and from an ocular point of view it's definitely not going to have any impact.

 

[DaveC426913]

I really don't know where this "speed of separation greater than c" is coming from. IMO, it has nothing to do with the OP's question and simply confuses the issue.

 

[lightarrow]

This formula is fine(ish) for calculating relative observed speeds but not much use for un-observable speeds.
This following text comes from another post on this forum which I thought illustrates the point I am trying to make.

Don't confuse what can be called the "3rd party separation rate" with relative speed.

Imagine this situation: Spaceship A is 1 light year to the west of me and travels at a speed of 0.9999c towards me; Spaceship B is 1 light year to the east of me and travels at a speed of 0.9999c towards me. What's their distance apart according to me? 2 light-years. When do they reach me? In about 1 year. So the separation distance decreases at a rate of about 2c according to me.

Of course the relative speed of spaceship B as measured by spaceship A is still less than c.

Look, I can make mathematical symbols do what they wants, I can add c as many times as I want, but it doesn't produce a *physical* velocity. Physics IS NOT mathematics. You only have to understand this. If you want to talk about physics, then you have to talk about things which have a physical meaning, not something else, otherwise you can post on the mathematics forum.

 

[DaveC426913]

Look, I can make mathematical symbols do what they wants, I can add c as many times as I want, but it doesn't produce a *physical* velocity. Physics IS NOT mathematics. You only have to understand this. If you want to talk about physics, then you have to talk about things which have a physical meaning, not something else, otherwise you can post on the mathematics forum.

What part exactly of what he is saying is physically nonsensical?

 

[Dale]

Physics IS NOT mathematics.

This is, on its face, a true statement. However, IF the universe behaves logically THEN the universe can be described in mathematical terms. Math is the language of logic.

Thus far, it appears that the universe is logical, and the mathematical rules that we are describing in this thread are not arbitrary, but are a way of describing the laws of nature clearly and concisely. You cannot dismiss the role of mathematics in physics, without math all of the elegance and logic of physics would be lost.

 

[russ_watters]

Now if we go for these two spaceships going passed the Earth and continuing on their way they will not be able to see each other although we still will.

That's wrong. Two space ships moving away from each other, each at .999 C, wrt a 3rd observer, will be able to see each other. Since their speed wrt each other is below C, a light beam sent from one to the other will catch the other, bounce off, and go back to the first ship.

Nor does even going faster than C necessarily prevent an object from being seen - as long as the light isn't coming from you. We see galaxies that appear to be receeding from us at greater than C.

 

[Doc Al]

This formula is fine(ish) for calculating relative observed speeds but not much use for un-observable speeds.

What do you mean by unobservable speed? Give an example.

This following text comes from another post on this forum which I thought illustrates the point I am trying to make.

Don't confuse what can be called the "3rd party separation rate" with relative speed.

Imagine this situation: Spaceship A is 1 light year to the west of me and travels at a speed of 0.9999c towards me; Spaceship B is 1 light year to the east of me and travels at a speed of 0.9999c towards me. What's their distance apart according to me? 2 light-years. When do they reach me? In about 1 year. So the separation distance decreases at a rate of about 2c according to me.

Of course the relative speed of spaceship B as measured by spaceship A is still less than c.

That brilliant quote about "3rd party separation rate" is from a post of mine. Reread it carefully, especially the last line.

Now if we go for these two spaceships going passed the Earth and continuing on their way they will not be able to see each other although we still will.

Why can't they see each other?

Going back to the original post on this thread. Noone on Earth will see the pilot walking towards the front of the spaceship assuming his 'rate of separation' from us exceeds c. We will see him again when his sits down at the controls.

Ah... I think I see what you're doing. You are misinterpreting that 'rate of separation'. As it would apply to this example, somebody at rest with respect to the ship would observe the Earth moving one way at some speed close to light speed and the pilot moving at some speed in the opposite direction. (For fun, let's have the pilot move through the ship at half the speed of light.) So, according to the stationary observers in the ship (the "third party"), the Earth and the Pilot are separating at a rate of about 1.5 c. So what? With respect to everyone, the pilot still moves at a speed less than the speed of light. In particular, the people on Earth will see the pilot moving at some speed less than c. (Very close to c, since the ship is almost traveling at light speed.)

 

[matheinste]

Hello Nickleodeon.

Quote:-

---I don't disagree with this except for the fact that after the speed of separation is greater than c the wavelength becomes too great to do any measuring and from an ocular point of view it's definitely not going to have any impact. ---

Note that although the speed of separation viewed by a third obserrver can approach 2c, each of the two seperating objects sees the other moving at less than c. No-one in SR sees ( measures is a better word in these circumstances as seeing can be mistaken for the visual effect ) anyone else moving at greater than c. Under these circumstances the remark about the wavelength is irrelevant.

I see Doc all beat me too it and put it better than i did.

Matheinste.

Matheinste.

 

[Nickelodeon]

What do you mean by unobservable speed? Give an example.

Just guessing here but dark matter would probably fall into this category as would whatever is going on within the event horizon of a black hole.

That brilliant quote about "3rd party separation rate" is from a post of mine. Reread it carefully, especially the last line.

Yes it was your quote - I do agree it was pretty good

Why can't they see each other?

I think it's red-shifting taken to extreme. After c the wavelength goes infinite. Here again the spaceship has to be going away from the observer not across his line of vision.

 

[Doc Al]

Just guessing here but dark matter would probably fall into this category as would whatever is going on within the event horizon of a black hole.

How is that relevant to this thread?

I think it's red-shifting taken to extreme. After c the wavelength goes infinite. Here again the spaceship has to be going away from the observer not across his line of vision.

Again, it seems like you are still thinking that somehow the pilot's speed with respect to the Earth is greater than c. Not so.