Muon321 Posted March 10, 2013 Share Posted March 10, 2013 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? Link to comment Share on other sites More sharing options...
elfmotat Posted March 10, 2013 Share Posted March 10, 2013 (edited) This is just algebra: [math]\frac{kx}{x}=k~~ \text{if}~x\neq 0[/math] It's undefined if x=0. Edited March 10, 2013 by elfmotat Link to comment Share on other sites More sharing options...
timo Posted March 10, 2013 Share Posted March 10, 2013 (edited) 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, 2013 by timo Link to comment Share on other sites More sharing options...
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