# CERN, speed of light

1. May 5, 2008

### Rikendogenz

Please correct me where I'm wrong,
Our solar system is moving around our galaxy at 250km/s and CERN can accelerate a particle to 99.99% the speed of light. If you are observing a particle going 99.99% of c from outside our galaxy and it happens to be moving the same direction of our solar system, the total speed would be greater than c.... Am I looking at this the wrong way?

2. May 5, 2008

### Janus

Staff Emeritus
You have to use the Relativistic formula for the addtion of velocities:

$$w = \frac{u+v}{1+\frac{uv}{c^2}}$$

This will always give an answer of less than c.

BTW, this is the correct formula to use when adding any velocites, But when u and v are small compared to c, the answer comes out to be so close to the answer you get when you just use $w=u+v$, you can use the simpler formula as long as your answer doesn't have to be too exact.

3. May 5, 2008

### Hootenanny

Staff Emeritus
Hi Rikendogenz and welcome to PF,

In special relativity speed is not an additive quantity. For example in Euclidean relativity (every day relativity), if your running left at 5 m/s and I am running right at 5 m/s, then my speed relative to you is 5 + 5 = 10 m/s. However, in special relativity this isn't the case one must use Lorentz transformations to determine the relatvie velocity of two moving bodies.

Edit: I see Janus has snook in before me

4. May 5, 2008

### JTankers

Head on photon 'collissions' have net 2C as observed from Stationary

From a stationary reference frame (Earth is close to a stationary frame... read special relativity, Einstein talks of the either and preferred reference frames as Isaac Newton did before him, other reference frames are primarily for time dilation...)

the following was a response I made to a similar question...

The Large Hadron Collider will collide proton to proton (and sometimes proton to anti-proton) in head on collisions with each set of particles travelling at 99.9999991% of the speed of light. The net collision speed is additive (same calculation as head-on car collisions). The kinetic energy per particle is determined by how close to the speed of light each particle travels. (That is why cosmic rays can have more energy even when the collision is moving to stationary rather than head on at same speed. The difference is how the energy is focused. Head-on collisions of same mass, same speed, exact opposite vector and center mass impact will focus the energy to a single point).

The particles will be guided around the tunnel by more than 1,600 superpowerful, cylinder-shaped ELECTROMAGNETS, some of which weigh more than 30 tons. The protons will zoom around the ring up to 11,245 times per second, reaching 99.9999991% of the speed of light.

At four points in the ring, magnets will push the beams together, causing up to 600 million PROTON COLLISIONS per second. If all goes as planned, these high-speed, high-energy crashes will create bursts of rare forces and particles that haven’t been seen since the big bang 13.7 billion years ago.

Four huge PARTICLE DETECTORS — the biggest, ATLAS, is 150 feet long, 82 feet high, and has more than 100 million sensors — will track and measure the particles at each collision. Filters will discard all but the 100 most interesting crashes per second. This will still produce enough data to fill a 12-mile-high stack of CDs per year.

JTankers

Last edited by a moderator: May 5, 2008
5. May 5, 2008

### RandallB

So where is the "net 2c" speed of anything?
Are you really going to add 99% + 99% for a speed faster than 100% of light "c"!
Did you even read post #4 by Janus on how to add speeds.

You are new here; you need to actually read what is expected of you in the must read first sticky posts.

6. Dec 7, 2011

### Prof Niemand

Strictly speaking, SR applies only in the absence of a gravitational field. In GR “c” is not the ultimate speed of either light or material objects, when a gravitational field is present.

In GR, the ultimate speed is determined only by the metric tensor. Following Max Born, if for simplicity we imagine a 2-D subset (x,t) of the 4-D spacetime continuum, and assume the off-diagonal elements of the metric tensor vanish, then the light lines are given by:

ds^2 = g11*dx^2 + g44*dt^2 = 0 --> ultimate speed = dx/dt = SQRT(-g44/g11).

In the flat Minkowski spacetime of SR, g11 = 1 & g44 = -c^2, so we have:

ultimate speed = SQRT[-(-c^2)/1] = c, as expected.

But in the curved Riemannian spacetime that exists when a gravitational field is present (or if you are using a non-inertial frame of reference?), the values of g11 & g44 could in principle be any real numbers, thus placing no theoretical upper limit on the ultimate speed in GR.

7. Dec 7, 2011

### PAllen

But that's strictly coordinate speed, not what anyone would measure. If you take the tangent vector of a null path expressed in the orthonormal basis at a point on timelike world line, you always get c exactly. This is the mathematical expression of 'local vaccuum speed of light is always c'. It is also required by the fact that semi-Riemannian geometry has flat Minkowski space as a tangent space at every point.