Travelling at the speed of light

[Ronnie67]

If we travelled at the speed of light and fired a gun, would the bullet slowley emerged from the barel or would it speed up?

[JamesF]

Neither. It would turn into a blue whale and a bowl of geraniums.

James

[ollypenrice]

Objects with mass cannot attain the speed of light so the question can't be answered.

But if an imaginary observer riding on a photon and approching an oncoming photon measures the oncoming photon's speed of approach he does not find that it is twice c but merely c.

On the other hand if Jeremy Clarckson and James May, both doing 100mph on a collision course towards each other, measure the approach speed of the other they will measure 200mph and, with luck, will be too busy calculating to take evasive action.

Olly

[DirkSteele]

If we travelled at the speed of light and fired a gun, would the bullet slowley emerged from the barel or would it speed up?

Funny, that is a similar to where Einstein's mind was when he started conducting the thought experiments that led to the development of Special Relativity.

I would clarify one point made above. Though it is not possible for an object with mass to reach the speed of light, this only applies to the speed of light in a vacuum (a universal constant and maximum speed limit of the universe, which is called "c"), it is possible for mass to exceed the speed of light travelling through a medium where the local speed of light can be far less than c. For example, the speed of propagation of light in water is 0.75c. It is possible to acclerate particles of mass beyond this (though still less than c). The famous consequence of this is Cherenkov radiation, which is what gives nuclear reactors their characteristic blue glow.

[Stu]

Slight correction to Olly's second example. I believe there is another fundamental law of physics which prevents James 'Captain Slow' May from exceeding 50 mph at sea level but otherwise it was correct

Stu

[Mikejh]

c+c=c. Sadly I never could visualise this concept.

[Treeden]

... and, with luck, will be too busy calculating to take evasive action.

+1

[Auntystatic]

so then two objects travelling toward each other at 0.5c = c (combined approach speed) ?

[Treeden]

so then two objects travelling toward each other at 0.5c = c (combined approach speed) ?

That seems reasonable to me. I don't think either would incur Mr Einstein's displeasure?

[George Jones]

so then two objects travelling toward each other at 0.5c = c (combined approach speed) ?

Nope. The combined speed is (4/5) c.

Speeds v1 and v2 are combined using

(v1 + v2)/(1 + v1*v2/c²)

[Treeden]

Nope. The combined speed is (4/5) c.

Speeds v1 and v2 are combined using

(v1 + v2)/(1 + v1*v2/c²)

I think I should wait an hour or so before I offer an opinion in here!

Thanks for that George, but please could you explain (in simple terms) why they don't have a simple combined closing velocity, the way two aircraft would that are on opposite headings in the same piece of sky?

Or would the formula still apply but the velocities are so small that it would barely make a difference?

Thanks

Rob

[JamesF]

I'd hazard a guess that at 0.5c, relativity is sticking its oar in?

James

[Treeden]

I'd hazard a guess that at 0.5c, relativity is sticking its oar in?

James

Hmm, I'm sure your right James, but I'm not quite there yet. Does that mean that if they are at opposite ends of the universe but still heading towards each other, then that would be ok?

[George Jones]

Thanks for that George, but please could you explain (in simple terms) why they don't have a simple combined closing velocity, the way two aircraft would that are on opposite headings in the same piece of sky?

I'd hazard a guess that at 0.5c, relativity is sticking its oar in?

Yup.

Suppose I observe body 1 coming at me from the left with speed v1 and body 2 coming at me from the right with speed v2. Speed = distance/time, and speeds v1 and v2 are computed using my time and distance. Closing speed V is, for example, the speed of body 2 with respect to body 1. Speed V is computed using body 1's distance and time. The theory of relativity says that time and space are relative, so distance and time for me and for body 1 are different, which means that V isn't simply v1 + v2.

Or would the formula still apply but the velocities are so small that it would barely make a difference?

Yes, when v1 and and v2 are small (even aircraft speeds), v1*v2/c² is not measurable different from zero, and V is not measurably different from v1 + v2 unless extremely accurate lab equipment is used.

[Treeden]

Suppose I observe body 1 coming at me from the left with speed v1 and body 2 coming at me from the right with speed v2. Speed = distance/time, and speeds v1 and v2 are computed using my time and distance. Closing speed V is, for example, the speed of body 2 with respect to body 1. Speed V is computed using body 1's distance and time. The theory of relativity says that time and space are relative, so distance and time for me and for body 1 are different, which means that V isn't simply v1 + v2.

Thanks George.

I understand can accept that relative to an observer (a third party) it would be so, but I'm struggling to see how each of the two individual bodies doesn't have the same relativity with regard to each other?

If you observe yourself walking directly towards a large mirror, would you not see (observe, calculate etc) that the distance is being reduced at a rate that is equal to twice the speed that you are walking at?

Or is that different?

[Mikejh]

Photon A and Photon B are travelling towards each other at SOL. Closing speed = 2xSOL but neither break the rule as they are "only" travelling at SOL? Physically not possible, but suppose you could put a torch on photon A and shine it towards Photon B-what then?

[JamesF]

I'm not sure that answering questions about situations that are physically impossible is particularly enlightening If you're doing something that's physically impossible, what physical laws have you broken to do so and in what respects do those same laws now apply or not apply to everything else?

James

[sonic190178]

Nope. The combined speed is (4/5) c.

Speeds v1 and v2 are combined using

(v1 + v2)/(1 + v1*v2/c²)

george if two spaceships where one million light years away and were traveling at half the speed of light then the closing speed would be the speed of light

as only the spaceships would be traveling the space that is in between them is not traveling at the speed of light but the closing distance is the speed of light

[Mikejh]

James you are right-let me ask in a different way. Train travels at 200mph. Headlamp on front of train. Light from train travels at SOL so without understanding the science you might expect the light to travel at SOL+speed of train - I am told it cannot-just don't understand why not!

[JulianO]

Photon A and Photon B are travelling towards each other at SOL. Closing speed = 2xSOL but neither break the rule as they are "only" travelling at SOL? Physically not possible, but suppose you could put a torch on photon A and shine it towards Photon B-what then?

As you say it cant be done in reality, but ignoring that.... its not that simple.

Let assume you're in a rocket going at nearly the speed of light. Firstly, how do you know you are, and that its just not things rushing towards you at nearly the speed of light? You can't tell, either view works. Its like sitting on a train in a station and the train next to you pulls out - its hard to tell which of you is moving - apart from the acceleration.

Anyway, you're in your rocket, either going at nearly the speed of light, or perhaps standing still and things moving around you. You shine a light from the nose, the light goes off at c. At least it does from where you're sitting in the rocket.

If your on another rocket heading towards the first, at nearly the speed of light, (or perhaps standing still with everything rushing towards you) you'll see the other rocket heading towards you and it's light.

How are you going to work out the closing speed? Probably bounce some light off the rocket and see how long it takes to come back, as that's the fastest thing you have at you're disposal. The light ray (radar maybe) leaves your rocket at c and bounces back.

It takes time T to come back, so you know how far away it is (s=cT), and its also blue shifted because it's approaching you, so that tells you velocity. Its going to come out like Georges formula above. Its like you're standing still and the other rocket is coming to you with velocity

(v + v)/(1 + v*v/c²) = 2v/(1+v²/c²)

assuming both rockets are travelling at v. If v is practically c, then this is roughly

2v/(1+1) = v

so the ship appears to be coming at about c.

What we normally do when adding velocities is wrong, if I fire a gun out of a car going 60mph, and the bullet is doing 100mph as it leaves the muzzle, then the combined velocity isn't really 160mph as you might think as a bystander. It's really

(100+60)/(1+100*60/c²)

which is

160/(1+6000/670,000,000²) = 160/1.00000000000001334143712707025197076955369942541917

= 159.99999999999786537005966878816370800989867816271256 mph

roughly

[JamesF]

The short answer is, I guess, that at such high velocities the normal Newtonian laws of motion and understanding of time and space on which we all base our every day intuition don't actually apply.

The speed of light is constant. If you're on the train, you will "see" the photons travelling away from the train at c. If you're standing outside on the station, you'll see the photos travelling away from the train at c. How can that be if the train is travelling at 200mph? Because when you're in motion relative to another observer, time passes at a different rate for each of you. And that different rate allows for the photons to appear to travel at the same speed in each of your cases.

I think I still do not find relativity trivial to get my brain around.

James

[JamesF]

If that's wrong. btw, please do point it out, someone

James

[mightyboo]

Wow the wife and i loved that cheers guys .

There's some lovely mud over here .