# Objects colliding at relativistic speeds combined energy?

LHC is currently capable of accelerating a batch of atoms faster than $0.5 \, c$.

What if we made a second LHC that intersects the original at, lets say the atlas chamber, but that runs in the opposite direction. Assuming that the batches actually front-collide (which is a long shot I know), isn't the energy (not the relative speed) involved in the collision higher than that in a batch hitting a static object?

• possible duplicate of Double light speed – John Rennie Aug 4 '15 at 19:09
• Hi Leonardo. The LHC collides protons moving at 0.999999991$c$ in one direction with protons moving at 0.999999991$c$ in the opposite direction. However as far as the protons are concerned the collision speed is still below $c$ because velocities don't simply add together when you get near the speed of light. The question I've linked as a duplicate explains what actually happens. – John Rennie Aug 4 '15 at 19:12
• @JohnRennie tks John i did not know that the beams were already in that setup. I do understand that the speed does not combine, but how about the total energy of the system? If the energy does not combine also, why do they bother rotating them to force a front collision? – Leonardo Aug 4 '15 at 19:51
• @Leonardo, I made that edit to help you use proper English grammar and spelling. This helps people understand your question and makes it more likely that you'll get an answer. Rolling back an edit like that is not a good idea. Good luck using this site. – DanielSank Aug 4 '15 at 20:17
• People, this question is not asking about getting relative speeds higher than $c$. Read the actual question! Leonardo is asking about the total energy in the collision. Y'all have itchy close-voting fingers. – DanielSank Aug 4 '15 at 20:24

isn't the energy (not the relative speed) involved in the collision higher than that in a batch hitting a static object?

Yes it is. It's double. Conservation of energy applies. You give some kinetic energy x to one object to make it move, and you give the same kinetic energy x to another object. When they collide head on, the collision energy is 2x. See the Wikipedia LHC article which says this:

"The protons will each have an energy of 7 TeV, giving a total collision energy of 14 TeV".

The relative speed is arguably double too, though IMHO it's better to refer to the closing speed. Again from the Wikipedia article:

"At this energy the protons have a Lorentz factor of about 7,500 and move at about 0.999999991 c, or about 3 metres per second slower than the speed of light".

One proton comes in this way → at 299,792,455 m/s, the other comes in this way ← at 299,792,455 m/s. Their closing speed is 599,584,910 m/s or just under 2c. Check out the closing speed section of the Wikipedia faster-than-light article. Note that the velocity addition formula:

$$s = {v+u \over 1+(vu/c^2)}$$

...tells you how fast you would measure another object approaching you, and this is always going to be less than c. If you were one of the protons, you might claim that the other proton was coming at you at 0.999999999999999959c, and you might call this the relative speed. But make no mistake, closing speeds can approach 2c. Just think about watching two spaceships setting out towards one another on a collision course from stars 4 light years apart. You would see them crash into one another after 2 years.