From the point of veiw of the first particle, space is warped so it would measure the speed of the second particle coming towards it at a speed less than c... probably something like 0.9998c, but it's too early for me to remember the equations. The speed you see each particle travelling at depends on where you are when you do the measurement: travelling with one of the particles, or off to one side and stationary with the surface of the Earth (which is not stationary in space anyway...)



The particles are atomic nuclei and/or protons, about 10^(-30)kg. The actual mass they reach will merely be "very heavy" since it depends on the power you can put into the collider system.

Hmm.

You to realise that while 'c' is the one absolute measure in the universe, everything else is relative.

Look at it this way, you have particle A and particle B. If you were a really really tiny person and were stood on particle A, particle B would be moving at very close to c relatively speaking but particle A wouldn't be moving at all relatively speaking, so their combined velocity from your perspective is still lower than c. The same in reverse if you were stood on particle B.

if you're looking from a third point in space, then their velocities relative to you would have to be worked out and summed according to standard physics. More than likely one particle would be moving towards you but the other one would be moving away from you, so the sum of their relative velocities would still end up being less than c.

This kind of stuff looks a bit 'fixed' to be honest but the maths does fall out right. It's actually impossible for two particles to have a combined relative velocity any greater than c.

of course, an engineer would observe that their relative velocity at the moment of impact is actually zero....

Jon

That's what I said

At the working energy of the LHV (7TeV), the particles weigh 10^-27 kg

There is a theory that light isn't actually constant and that the speed of light was faster at the dawn of the Universe.

What that has to do with cars I have no idea.

Light's what comes out of the front of cars when it's dark.

Slightly off topic, but two identical cars doing 70 in a head on collision (in isolation from other objects) is identical to a car doing 140 hitting a stationary car.

In the first case, both cars go from 70 to zero. In the latter case, momentum is conserved and after the collision both cars would be doing 70.

For any inertial observer, the change in energy and momentum of the system would be the same:

For two cars head on at speed v, the total conversion of kinetic energy into pain = 2 x ½Mv² = Mv²
For one car at speed 2v hitting a stationary car, loss of kinetic energy = ½M(2v)² - 2 x ½Mv² = Mv²

Note that hitting a car coming the other way at the same speed is the same as hitting an immovable stationary car: you decelerate from v to 0. That's part of why you shouldn't drive into Volvos full of depleted uranium, or trees.

I can't remember the equation, it's been years, but I'm sure there is one that relates to velocities mass and light that is formulated so that any result can never be greater than the speed of light.

Vague and not very useful, I know.

I believe I speak for everyone when I say that before this thread, the theory of relativity was merely a vague concept to me, full of ingraspable equations and unsatisfactory explanations. Now, from the mountaintops of understanding, it's like looking out towards the glowing horizon of a clarified new world, full of excitement and branching possibilites. Thanks everyone *


*the above post may contain sarcasm.

Er. No.

(1/2)*(2^2) is 2, not 1. The second total is 2M*v^2.

The two scenarios you describe are different; you are not describing the same situation from two viewpoints (where conservation would apply and the totals would be equal).

In the first case, both the cars are moving with respect to the ground (by which their velocity v is defined). In the second case, only one car is moving with respect to the ground, at twice the speed (and with four times the energy) of one of the cars from the first situation. In case 2 there is more kinetic energy in the system, wherever you observe it from.



Edit: so driving at 140mph into a wall is worse than hitting head on at 2*70mph. You are twice as dead...

Check your maths, you'll see I'm quite correct. Of course ½ x 2² does not equal 1, and I never said any such thing.

Let me break it down for you:

½M(2v)² (the energy of one vehicle at 2v prior to collision) - 2 x ½Mv² (the energy of two vehicles at v after the collision) =

= ½M*2²*v² - Mv²

= ½M*4*v² - Mv²

= 2Mv² - Mv²

= Mv²

Which equals the combined energy of 2 x vehicles with energy ½Mv² as I originally stated.

You can nay change the laws of physics


You're missing the point of relativity. If you shift your point of view from the ground to the driver, then you should understand better. The accident would feel precisely the same if your relative impact speed was 140mph, while traveling at 10,000 mph through space relative to "the ground".

The truth is, There is no ground. All inertial reference points are equivalent.

*inserts foot in mouth*

You got maths owned dude.

LOL.

I should point out that a real traffic accident involving a car doing 140mph is indeed more serious, because after the initial impact there is almost certain to be a secondary impact with something on "the ground". This is analogous to the 70-70 head on collision, followed by "the ground" running into you at 70mph. Here on earth, "the ground" is quite big and should generally not be excluded from calculations it may impact upon.

However, as a metaphor for a scientific experiment in a particle accelerator such things are not relevant. We're only interested in the energy of the initial collision.

Lol I'm siding with JJW009 on this one. Law of conservation of energy or something. We did the same experiment using little wooden blocks with wheels to represent cars, a piece of card in the top of each to trigger the light sensor, two light gates and two ramps.