Let's say that we had some equation in physics that was v(m+c)/v, which isn't real I am just using it, would you be able to simplify it to m+c? You shouldn't be able to because v for velocity has a possibility of being 0, correct? You can't simplify if you would be cancelling out a variable that could be zero, is this true?

This is just algebra:

 

[math]\frac{kx}{x}=k~~ \text{if}~x\neq 0[/math]

 

It's undefined if x=0.

Edited March 10, 201312 yr by elfmotat

Let's say that we had some equation in physics that was v(m+c)/v, which isn't real I am just using it, would you be able to simplify it to m+c? You shouldn't be able to because v for velocity has a possibility of being 0, correct? You can't simplify if you would be cancelling out a variable that could be zero, is this true?

There is no clear rule about it, in your example. The original expression obviously does not apply for v=0. In all other cases, the 2nd expression is equal to the first one. So in all cases where the 1st expression applies (*), the 2nd expression is identical, and therefore a valid simplification. On the purely mathematical level, the only difference is that the 2nd expression doesn't explicitly show that the equation isn't supposed to hold true for v=0.

 

If you switch from this out-of-context view of your example to real physics, then things get even more interesting: Assuming that the physical quantity the term represents does not run into some magical undefinedness at v=0 but is somewhat continuous there (**), then the 2nd expression already includes the limit that gives a proper description of what happens at v=0. In this case, the 2nd term may be more appropriate than the 1st.

 

Bottom line: Contrary to widespread public opinion, physics is not math.

 

 

 

(*) Strictly speaking: where it formally applies. Physics equations usually come with a lot more restrictions about the applicability, that are usually not mentioned explicitly but assumed to be understood by the reader (who usually is assumed to be a proper physicist or engineer).

 

(**) Note that a physical observable with a discontinuity at v=0 but well-defined behaviour arbitrary close to v=0 would be a strange beast: If an infinitely small deviation from v=0 leads to a completely different behaviour of the variable, then considering you cannot measure anything to an arbitrary degree of accuracy your model to describe nature may be useless, anyway - depending on the context.

Edited March 10, 201312 yr by timo