What does "m0 → 0" mean in physics

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SUMMARY

The expression "m0 → 0" in physics typically refers to the concept of stationary mass, where "m0" denotes the rest mass of a particle. In the context of special relativity, "m0" is considered an invariant quantity. The notation "m0 → 0" may indicate a scenario where the mass of a component in a physical system is reduced, affecting the system's behavior. However, the interpretation of this expression heavily relies on the specific context in which it is used.

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The symbol "m0" means stationary mass, but what does "m0 → 0" mean in physics?
 
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As with just about all such questions, it depends on the context... so you'll have to tell us where you came across this expression before we can help you with it.
 
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@ Nugatory:
I encountered such a question in another site last month. But now piggy is here and reluctant to link those context. Ah, can we go on with the topic in a way as below?

In math we can consider that “m0” is just a mathematical symbol. We can carry out such mathematical operations as “m0 → ∞ ” or “m0 → 0”.
But in physics, for a certain particle, (if no other miscellaneous factors meddle in), m0 is an invariable.
 
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If you refuse to tell us what this is all about, how do you expect us to help you?
 
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htam9876 said:
But in physics, for a certain particle, (if no other miscellaneous factors meddle in), m0 is an invariable.
It is true that ##m_0## is often used to represent the rest mass of something, especially in the context of special relativity, but that's not the only way it’s used. And even in that specific context, the notation ##m_0\rightarrow 0## might be part of a discussion of how a particular physical system behaves if we replace one part of with something less massive... or it might be something completely different.

So there is no answer to your question unless and until you supply that context.
 
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You can be sad about my post all you want. But it is nevertheless true that if you refuse to tell us what this is all about, we can't help you.

Your "case" has neither an m0, nor a zero, nor a discernible point.
 
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Vanadium 50 said:
Your "case" has neither an m0, nor a zero, nor a discernible point.
With that being the case, the thread is now closed.
 

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