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Partial Derivatives in Physics

Jonathan Enders

September 22, 2009

Partial derivatives lead to a lot of confusion for many who start learning physics. But they are excessivelyencountered: Lagranges equation and thermodynamical potentials are just a few but prominent exam-ples. On a close look one can come to the following conclusion to why partial derivatives are somewhattroubling.

Many physicists do not make a distinction between a function y(x) and a variable y. They tend totreat every equation as a relation that exists among the variables present in the equation. While thisis perfectly fine and helpful when dealing with total derivatives it will cause trouble when dealing withpartial derivatives.

Example:

The following relation is given: x= t +

When calculating the total derivative dxdt

nothing can go wrong. You can simply plug in the right side of

the relation for x and that will give you the total derivative dxdt

= d(t+)dt

= d(t)dt

+ d()dt

= 1 + ddt

.

In total derivatives the variable(s) in the nominator expression can be replaced by everything that isequal

When calculating the partial derivative xt

you could be tempted to do the same as above, but this is

wrong. The result of xt

is really 0. Only xx

and tt

are 1. The source of this confusion is that many

times xt

is written when it should be x(t)t

. x is a variable and x(t) is a function.

You could define x(t) as being t+ . Then the expression x(t)t

can be evaluated, because t+ canbe inserted for x(t). Now an important property of a functionx(t) as opposed to a variable x becomes

clear. x(t) is defined as being t+ which is equal to x. But if you would plug in x for x(t) in thepartial derivative you would get a wrong result x(t)

t = x

t= 0 instead of x(t)

t = (t+)

t = t

t+

t = 1.

This shows that there is a difference between defining a function x(t) := t+ 1 and setting a functionx(t) = t + 1. A function can be equal to a lot of things x(t) = t+ 1 = x but can only be defined once(x(t) := t + 1 or x(t) := x).

In partial derivativesonlythe function(s) in the nominator expression can be replacedonlyby the(ir)definition(s)

Now many of the famous equations in physics contain partial derivatives. For example the Lagrange

equation: ddt

L(q,q)q

L(q,q)q

= 0

Note that L(q, q) is a function not a variable. An equation like that would be completely useless ifthe function in the partial derivative wasnt further specified. We know that L(q, q) is defined asL(q, q) := T V. Note here again that the previous expression is wrong because T and V are reallyfunctions and not variables. So to make everything very clear the expression has to look like this:L(q, q) :=T(q) V(q).This is all we know. We have to define the functions T(q) and V(q) ourselves meaning we have to findan expression for the kinetic energy that is only dependent on qand we have to find an expression forthe potential that only depends on q. That is essentially the step that lets the Lagrange equation knowwhat you want it to solve. The rest is calculation.

Another really head breaking example that is incredibly worth noting is the one of the thermodynamicalpotentials. Physicists have developed a funny notation in order to make sure that the right functions areinserted in the equations. It looks like this:T = (U

S) |V,N

This is the partial derivative of the internal energy with respect to the entropy while holding the volumeand the number of particles constant. What this actually means is that the function for the internalenergy you plug in ought to depend only on the variables entropy (S), volume (V) and particle count(N). Thus, they also could have just agreed on the following notation:

T = (U(S,V,N)S

)