Is breaking the speed of light limit easier going straight forward rather than circular?

[gnoitall]

@Pasta

Menlo Park, California, is rather hilly, but the beamline happily punches through the ground in a straight line. The curvature of the earth introduces no additional complexity. In fact, it would be harder to follow the curvature of the earth than to go in a straight line. To your broader point, the curvature of the earth isn’t a deciding factor between a linear or circular accelerator.

Agreed. A linear PA doesn’t do any kind of beam steering (the magnets that force the particles to circle) in the main beampath, so it has to be literally laser-beam straight. That also puts a practical limitation on how long one can be, since it becomes a question of “how long a perfectly straight tube through the earth can I drill?”

[Pasta]

@gnoitall

Agreed. A linear PA doesn’t do any kind of beam steering (the magnets that force the particles to circle) in the main beampath, so it has to be literally laser-beam straight. That also puts a practical limitation on how long one can be, since it becomes a question of “how long a perfectly straight tube through the earth can I drill?”

I guess I wouldn’t call that a “practical limitation”, is all. Even a 100 km straight line only needs a trivial depth of a couple hundred meters, and that’s already as long a beamline as would be limited by other practical considerations. I concur that drilling, say, 1000 km in a straight line is a challenge, but that doesn’t bear on accelerator design in any practical way.

[Riemann]

@wolfpup

True, but in one particular type of circular motion, namely, an orbit around a center of gravity, no acceleration is detectable in the frame of reference of the orbiting body. ISTM that the consequence of this is that the same Lorentz transformation applies here as in linear motion, as seen here in the time correction necessary for GPS satellites due to orbital velocity. The effect of being further out from the earth’s gravity well is a different correction with the opposite sign applied independently according to general relativity.

Acceleration due to gravity is not sensed by anything following a GR geodesic (in free fall). But that doesn’t mean it’s an inertial reference frame - I don’t see how your claimed consequence follows.

[wolfpup]

@Riemann

Acceleration due to gravity is not sensed by anything following a GR geodesic (in free fall). But that doesn’t mean it’s an inertial reference frame - I don’t see how your claimed consequence follows.

Right, it doesn’t mean that it is, but is it equivalent to an inertial reference frame? It would seem to be a corollary of the equivalence principle – which states in essence that you cannot distinguish between acceleration due to gravity and acceleration due to a propulsive force – that you also cannot distinguish between an inertial reference frame and one that is in free fall.

My supposition may be wrong, but it seems to me that based on the above assumptions, for an orbiting object in which no acceleration can be detected, one can regard its scalar quantity “speed” as exactly equivalent to the vector “velocity” for a linearly moving object for the purpose of calculating the Lorentz transformation for time dilation due to velocity.

[komolono]

Any chance of getting remarkably more velocity if one adds a sling(linear tube) to CERN LHC?

[Francis_Vaughan]

@komolono

Any chance of getting remarkably more velocity if one adds a sling(linear tube) to CERN LHC?

Not really. Speeds don’t add like that. As I wrote above - you need to think energy, not speed. The LHC already runs at 0.9999999896 c You can run faster than the gap between the particle speed and c. But that doesn’t say that you can bridge that gap.

There are lots of complicating matters, but even if we are able to use all the energy imparted by an injecting accelerator to particles, and imagine that we transport SLAC to Geneva and aim it into the LHC* (this isn’t sensible or useful, for all sorts of reasons, but no matter, just roll with it) and we somehow manage to add the energies from the two, you add SLAC’s 90GeV to the 6.5TeV of the LHC, and, well you will still only have a tiny smidgin more Call it 6.59TeV.

*or the reverse, it doesn’t matter here

0.9999999896 c = 299792.45487 km/s at 6.5TeV
to
0.9999999898 = 299792.45496 km/s at 6.59TeV

That is a difference of 0.09m/s You can crawl along the ground faster than that.

(Numbers based on a quoted Lorentz factor of 6,930 for a claimed 6.5TeV. I’m too lazy to calculate the whole thing out, but the final answer is going to be pretty close.)

[Pasta]

@komolono

Any chance of getting remarkably more velocity if one adds a sling(linear tube) to CERN LHC?

As mentioned above, the speed limit at the LHC is due to the bending magnets. The LHC can already accelerate the particles more. It just can’t turn them around the bends if they go any faster.