# What exactly is an invariant quantity?

I have a bit of confusion regarding an invariant quantity. It is something which doesn't change on switching from one inertial frame to other like $\Delta$$\mu$$J$$\mu is an invariant. I read transformation of fields from Feynman Volume 2 and noticed that E$$x$$'=E$$x$. So can I call $E$$x an invariant? (E$$x$ is the field due to a charged particle moving along X-axis).

Also I have other confusion, if I choose a physically identifiable point (say any point on a wall perpendicular to the moving charge), does the numerical value of Electric field remains same in both the frames and $E$ transforms in such a way to give same numerical value of $E$ or both observers obtain completely different numerical values as well. And if numerical value changes, wouldn't it mean that charged particle experiences a different force in two inertial frames?

EDIT: To clarify, does an invariant mean that the function returns same value at a point in all inertial frames or it means that the functional form remains same (say that something falls as $\frac{1}{x^2}$ in all frames where $x$ corresponds to that frame's $x$).

"Invariance" by itself simply means "not changing". What it means precisely is, like most words, context dependent. A quantity might be invariant in one sense and not invariant in another.

Most of the time, when physicists talk about invariance, they are talking about "frame-invariance". This would mean that the quantity is invariant (not changing) under arbitrary (or a special sub-set of) transformations in reference frames. Caution should be made here though that what set of transformations of reference frames is being referred to is again context dependent. Galilean invariance vs Poincare invariance are different! In the modern world, the set of frame invariant objects would generally consist of the class of geometrical objects called "tensors" (of which "vectors" - strictly 4-vectors for SR/GR applications - are a subset).

Now, with the advent of SR/GR we found that the electric field is NOT a tensor object, and so generally we don't speak of the electric field as an "invariant". The electric field is in fact 3 components of an anti-symmetrical rank 2 tensor called the "Electromagnetic Field Tensor" (the other 3 independent components of this tensor is the magnetic field). The invariant object is this tensor (think - tensor is like the physical arrow that represents a vector), it's components (think, how do we express the components of a vector given a coordinate system) are subject to change depending on what coordinate transformations you are applying.

So, finally to your question. Just because $E_x=E_x'$ is true for a boost in the x-direction, does not mean this holds if I apply any other Lorentz transformation. If I boost in the y-direction, or if I simply rotate my coordinate system, it will certainly not be true that $E_x=E_x'$. In this sense, we usually do NOT say that $E_x$ is "invariant". We can say "$E_x$ is invariant to a boost in the x-direction", but we MUST mention the rest of that sentence "to a boost in the x-direction." If you say "$E_x$ is an invariant" then you are only serving to confuse everybody you are talking with.

It is generally better to think of invariance in terms of the geometrical object (tensors/vectors) rather than their components. Thinking in terms of specific components will only serve to confuse.

• What about "invariance" of the Lagrange function upon generalized coordinate transformations, or the Hamilton function upon canonical transformations? They do no seem to be "tensors". – freecharly Mar 28 '18 at 19:24
• Hence why I said "generally". And I also specified that the meaning of the word "invariant" is dependent on context. I'm not sure what your point is here. My answer should not serve as a reference to all possible uses of "invariance" in physics, but only to the use that is relevant to the OP's question. – enumaris Mar 28 '18 at 19:42
• @enumeraris - The OP is asking about the the value of functions. Also, SR posits the invariance of physical laws in inertial systems. – freecharly Mar 28 '18 at 21:27
• So my answer is wrong or missing something? You are free to provide a better answer yourself. – enumaris Mar 28 '18 at 21:36
• Not wrong, but it would be surely helpful for the OP if you also addressed these points. – freecharly Mar 28 '18 at 21:46