How to understand the true meaning of momentarily comoving reference frame? My understanding is that the comoving frame is defined as: at a certain moment and point, a comoving frame is the inertial frame which has the same velocity as the local particle, or maybe fluid element, which is usually considered accelerating.
This definition is clear, but somehow I am just not satisfied and think this definition does not fully serve the name "comoving". In my opinion, the purpose to introduce this concept is to simplify some derivation. Specifically, we expect an observer in the comoving frame should experience exactly the same physics as if one would experience when the fluid element keeps static (at least in that instant moment). But somehow I think the two observers can only agree on the physics when the comoving frame is also accelerating with the same acceleration.
I apologize that I can not well explain why I believe so since my logic is not well organized yet, probably because that, I think they can agree on the information about forces only after two frames have the same acceleration. Hope someone can help me understand this comoving frame concept, this really confuses me. Thanks in advance.
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I appreciate you all answering this ambiguous question, sorry I can not describe better. Maybe let me try another way, a typical application of the comoving frame is to derive the stress-energy tensor. Usually we start in a frame where the fluid is constantly static, then by arguments, we write down the simple stress-energy tensor in this frame, say it is $T_{\rm static}$. At last we transform $T_{\rm static}$ back to any frame we want.
Now imagine we have a field of flowing fluid, and the velocity field is not uniform or static, fluid elements are even accelerating. If I sit on a fluid element, what is the stress-energy tensor I will observe? Will it be $T_{\rm static}$? because in my vision this element is really always static. If I indeed observe $T_{\rm static}$, then why someone in an inertial frame (although comoving) should see the same thing as I do? since I thought we may have different opinions on forces or that sort of thing.
 A: 
this definition does not fully serve the name "comoving"

It is not merely “comoving” but “momentarily comoving”. You cannot neglect the word “momentarily”. It implies that the instantaneous velocity of the object is zero, not that the velocity is always zero.
The reason to include the word “momentarily” is because you can mathematically guarantee that there always exists a momentarily comoving inertial frame for any event on any timelike worldline.
You cannot make the same guarantee for frames that are comoving over a finite duration. Such frames are not generally inertial unless the object is inertial.
A: Your definition of a comoving observer is definitely correct. As another commenter has said, the "moment" matters: An observer is usually characterized by $(p,Z^a)$, where $p$ is an event on the worldline of the particle being observed, and $$Z^a=\left ( {\partial }/{\partial \tau}  \right ) ^a$$ is the 4-velocity of the observer. Notice the 4-velocity is nothing but a unit timelike tangent vector at $p$. If at $p$ the observer's 4-velocity equals the particle's 4-velocity $U^a$ (on a spacetime diagram the two worldlines are tangent at $p$), we should call such an observer an instantaneous comoving observer of the particle at $p$. Then, the timelike geodesic determined by $Z^a$ at $p$ (a straight line through $p$ in the direction of $Z^a$) should be called an instantaneous comoving inertial observer of the particle at $p$, a special case of an instantaneous comoving observer. Now, this inertial observer can be used to define an instantaneous inertial frame at $p$. The concept of an instantaneous observer amounts to a local measurement, which the observer can perform on his worldline. As for if the comoving observer experience the same physics as the particle being observed, globally, I don't think they would. A particle with a non-vanishing acceleration would experience fictitious forces, while the inertial observer who is only instantaneously comoving at a single point $p$ would never, and you could produce things like the "twins paradox". For example, some of the Christoffel symbols are non-vanishing in the Rindler coordinates of a particle undergoing constant proper acceleration, and so a pseudo force $f=-ma$ arises. Am I getting your question right?
To give a simple example for comoving observers, the components of the particle's 4-acceleration $A^a=U^b\nabla_b U^a$ are $$A^0=\gamma^4\mathbf{u} \cdot \mathbf{a},\:\:A^i=\gamma^2 a^i+\gamma^4u^i(\mathbf{u} \cdot \mathbf{a}) $$ according to an arbitrary instantaneous observer at $p$ who may or may not be comoving. Assuming the observer is instantaneously comoving, then in the associated inertial coordinate system, where the particle's instantaneous 3-velocity vanishes, we have $A^a=\mathbf{a}^a$, which is quite convenient (the particle's 4-acceleration equals its 3-acceleration at $p$). A non-comoving observer would disagree with that, but they would agree that $A^a$ is an absolute object which acquires different components under their inertial coordinate system. This is quite unlike the pseudo force, which is not a tensor at all because we can make instantaneous inertial coordinate systems where the Christoffel symbols all vanish, and so $f$ also vanishes. To give another example, a comoving observer would measure the energy density of a perfect fluid as $\mu$ at some point $p$, but a non-comoving observer will not measure the same energy density. To the same comoving observer, the fluid's 3-dimensional stress tensor would take a very simple form, where its components satisfy $T_{ij}=P\delta_{ij}$, as convenient as it can be.
