If an object is moving at a significant (reativistic) velocity wrt a particular frame of reference then that frame will see it as length contracted in the direction of motion.

Not only "see" it but the object will actually. be. length contracted in that frame of reference.

(Hope I am correct so far)

In the frame of the object itself no such contraction is observed or experienced.

So my question is ,are there any objects where this lack of symmetry would be a problem?

Are some objects required to be perfectly symmetrical along all 3 physical axes so that they could not physically exist "squashed up" in another frame of reference ?

(I presume the answer must be "no".Was the question worth asking -or well put?)

1 minute ago, geordief said:

So my question is ,are there any objects where this lack of symmetry would be a problem?

Depends on what constitutes a problem.

1 minute ago, geordief said:

Are some objects required to be perfectly symmetrical along all 3 physical axes so that they could not physically exist "squashed up" in another frame of reference ?

I don’t know about “required” but nuclei in particle accelerators are “pancaked” when observed from the lab frame.

As far as not being able to exist, it’s sort of an ill-formed proposition. There is always some frame where it’s true, i.e. some fast moving particle exists, somewhere, and that can’t affect what happens in the object’s rest frame. All of the laws of physics are the same regardless of your (inertial) frame. They are all subject to the same transform.

Put another way, there is no object that inherently spherical, because the shape is relative.

14 minutes ago, swansont said:

Put another way, there is no object that inherently spherical, because the shape is relative.

Probably what was on my mind.

Is it correct to say that there is no force in the universe that would be capable of achieving the same deformation of a nucleus ? (in a rest frame ,if that could apply as a description)

Nothing that could conceivably ,as it were put the nucleus in a vice and squeeze it so that its shape was identical to relativistic length contracted?

49 minutes ago, geordief said:

Probably what was on my mind.

Is it correct to say that there is no force in the universe that would be capable of achieving the same deformation of a nucleus ? (in a rest frame ,if that could apply as a description)

Nothing that could conceivably ,as it were put the nucleus in a vice and squeeze it so that its shape was identical to relativistic length contracted?

I don’t know, but nuclei do deform. There’s the liquid drop model. An excited nucleus will oscillate like a blob of fluid. For a large nucleus, the two lobes get far enough apart that they don’t attract (the strong force has a limited range) but still repel owing to the Coulomb force, and you get fission.

It should be noted that objects are really four-dimensional, each point of the three-dimensional object being a worldline in four-dimensional spacetime, with the appearance of the object in three dimensions at a given time being the section of a three-dimensional slice through the object. In the inertial frame of the object itself, this three-dimensional slice is perpendicular to the worldlines of the object, but from another inertial frame of reference at motion relative to the object, the three-dimensional slice is oblique to the worldlines of the object. For example, an oblique slice through a cylinder is elliptical, with the length of the circular section being elongated in the direction of the obliqueness. However, due to the peculiarity of the geometry of spacetime, lengths of spatial sections are shortened rather than lengthened. So, given that an inertial object is always at rest in its own inertial frame of reference, and all other inertial frames of reference exist regardless of whether there are any observers in them, length contraction is not a real phenomenon. However, it becomes more real for an accelerated object (which is always accelerating upward in its own frame of reference) because the bottom of the object has a greater acceleration than the top of the object. Nevertheless, the natural length of the object in its own accelerated frame of reference will remain constant unless forces associated with the acceleration physically distort the object.

Edited November 22Nov 22 by KJW

10 hours ago, geordief said:

So my question is ,are there any objects where this lack of symmetry would be a problem

I think a good way to think about this is to consider length contraction and time dilation (which always go together) as a relationship between frames, not as something that “happens” to clocks and rulers. You can specify how frames are related by stating their relative speed, or you can state the gamma factor (ie time dilation/length contraction) - these are entirely equivalent.

So there is no physical force that squeezes rulers, any more than there would be if you observed your house from a distance and noted that it has gotten smaller; all that changes is how observer and observed relate to one another. In SR, that relationship is one in spacetime.

Just be careful to not confuse length contraction with a mere optical illusion though - the flattened nucleus in the particle accelerator really does physically behave like a flattened disk in the lab frame, it doesn’t just “look” flattened.

9 hours ago, geordief said:

force in the universe that would be capable of achieving the same deformation of a nucleus

Inside a neutron star?

But surely, if one uses a 4 coordinate system, and sets \(t=ict\), then the coordinate transformation

\(C \to C'\) is orthoganal i.e. metric-preserving?