Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recall the definition of potential energy: $$ U_x-U_{x_0} = -\int^x_{x_0}F_x(x')dx' $$

I've seen the integral definition of work, but not this - the thing I'm specifically interested in is the notation. I've never seen differential and prime notation intermingled like this. What does it mean? Would it have been possible to simply express it as:

$$ U_x-U_{x_0} = -\int^x_{x_0}F_x(x)dx $$ or is there an important reason for the given notation?

share|improve this question

migrated from physics.stackexchange.com Nov 19 '13 at 17:16

This question came from our site for active researchers, academics and students of physics.

1  
    
A dummy variable is something you put in place of the original variable so as to avoid confusion. I could have written $U_x-U_{x_0} = -\int^x_{x_0}F(q)dq$ and it would still mean the very same thing –  Pranav Hosangadi Nov 19 '13 at 1:01

3 Answers 3

up vote 4 down vote accepted

$x'$ is a so-call dummy variable. The integral could just as well have been

$$ U_x-U_{x_0} = -\int^x_{x_0}F_x(\tau)d\tau $$

without changing the result.

The variable $\tau$ is "integrated out" and the result is a function of $x$ with $x_0$ as a parameter.

See, for example, the Wolfram article for Dummy Variable:

A variable that appears in a calculation only as a placeholder and which disappears completely in the final result.

share|improve this answer

What you do it's a simple change of variables. There is no change in anything, since the integration variable is mute. In this case the primed variable doesn't mean a derivative, but just as another character to be used as variable.

share|improve this answer

To add to what has been said, the second way you've expressed the integral is perhaps more natural but not rigorous. What happens when you take the derivative of the RHS with respect to $x$? One is tempted to differentiate everything containing $x$, getting the wrong result, instead of applying the fundamental theorem of calculus. To be safe, we use a dummy variable $x'$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.