# What does "Solutions" mean for Maxwell & Einstein Equations?

#### louarnold

What does "Solutions" mean for Maxwell & Einstein Equations?
I am a retired engineer studying (not in school) Maxwell's and Einstein's work at a beginner level. The concept of "solutions" to their equations comes up from time to time, and this puzzles me.

So I ask: What does it mean to have a "solution" to one of these equations? What is being solved for and why? I understand this question in the context of simple algebraic equations, but not for more complex systems such as Maxwell's wave equations and those of General Relativity. Can someone explain at appropriate level for me? Can someone point me to an example of a solution with obvious rational for it?
Thanks,
Lou.

#### studiot

Hello, Lou, and welcome.

Neither Maxwell's equations nor Einstein's are simple algebraic equations.

They are both a (different) shorthand notation for a system of simultaneous equations.
Some of these are differential equations.

So the first question is do you understand this and the terms

Independant variable
Dependant variable
Function
Matrix
Differential equation
Tensor

?

You need to understand the first three for certain,
The second three can be overcome by writing out all the equations in fukll, not as shorthand.
Engineers know that you cannot avoid the full vwerison when you put numbers in so they tend to work in full.
Physicists often use the shorthand version.

#### Pmb

PHF Hall of Fame
What does "Solutions" mean for Maxwell & Einstein Equations?
I am a retired engineer studying (not in school) Maxwell's and Einstein's work at a beginner level. The concept of "solutions" to their equations comes up from time to time, and this puzzles me.

So I ask: What does it mean to have a "solution" to one of these equations? What is being solved for and why? I understand this question in the context of simple algebraic equations, but not for more complex systems such as Maxwell's wave equations and those of General Relativity. Can someone explain at appropriate level for me? Can someone point me to an example of a solution with obvious rational for it?
Thanks,
Lou.
Hmmmmm. Is this what they call "The New Math"?

Something is said to be a solution of an equation when its inserted into the equation it yields an identity. The distribution of charge (for Maxwell's equations) and mass (for Einstein'/s equation, determines the value of the solution since there can be as many solutions as there are distributions of charge and mass.

For example: Pythagorean/s theorem is about right triangles. If you're not familiar with this then see where I provide a proof at

Pythagorean Theorem

In words it states that the sum of the square of the side and the square of the adjacent equals the square of the hypotenuse. This holds for all right triangles even when the lengths of the side and/or hypotenuse is different. In fact you can take any number and scale the adjacent and the opposite and it's still hold true. Think of this like a metaphor of a the equations you mention and what it means to be a solution and you'll be fine.

You can also find lecture notes on differential equation online and read the defintion there too. In fact that's where I got it from in "the olden days.'

#### studiot

PMB
Something is said to be a solution of an equation when its inserted into the equation it yields an identity

Consider the equation

x = 3

Insert +0 into the equation

x + 0 = 3

So by the above statement 0 is a solution of the equation?

#### Pmb

PHF Hall of Fame

Consider the equation

x = 3

Insert +0 into the equation

x + 0 = 3

So by the above statement 0 is a solution of the equation?
Of course not. That's not what it means to insert something into an equation. An equation is like a machine; put something in and get something out.

The person doing the work doesn't have access to those places. I meant when Q(q', b'', c'') = C is a differential equation then if a set is found such that q' = A, b'' = B and c''' = C are substituted into Q(q', b'', c'') = C and it still holds true then A, B and C is a solution set.;

As applied to EM: Suppose we take Gauss's law in a region where there is no charge at point P in space but there are charges elsewhere. Then

div E = 0

Where E is the electric field. Then its said that E is a solution to the equation (which is one of Maxwell's equations.

https://en.wikipedia.org/wiki/Ordinary_differential_equation

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#### studiot

So what is the solution to the equation in my example?

#### Pmb

PHF Hall of Fame
So what is the solution to the equation in my example?
I already explained, i.e. its meaningless to apply it to your example. Recall what I said

The person doing the work doesn't have access to those places. I meant when Q(q', b'', c'') = C is a differential equation then if a set is found such that q' = A, b'' = B and c''' = C are substituted into Q(q', b'', c'') = C and it still holds true then A, B and C is a solution set.;
I even gave you the precise definition in that Wikipedia page, specifically at
https://en.wikipedia.org/wiki/Ordinary_differential_equation#Solutions

You can also check out here: http://www.feynmanlectures.caltech.edu/II_20.html

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#### Pmb

PHF Hall of Fame

Consider the equation

x = 3

Insert +0 into the equation

x + 0 = 3

So by the above statement 0 is a solution of the equation?
Even after I read this a few times I still can't see how you got it wrong. The terminology was made clear by the examples.

Is this is serious question or a complaint that I didn't explain it well?

#### studiot

Even after I read this a few times I still can't see how you got it wrong. The terminology was made clear by the examples.

Is this is serious question or a complaint that I didn't explain it well?
I didn't get it wrong.

But I did take issue with your use of the word 'insert'

I note you later used the correct term substitution (the corresponding verb is of course 'to substutitute')

There is a world of difference between 'to insert and 'to substitute'.

When you insert you do not change anything that was already there, but you do add something that was not.

When you substitute you replace something(s) that was/were already there.

Failure to make this distinction leads to contorted thinking for example you surely can't be seriously claiming that asking for the solution to the equation x = 3 is meaningless?

Said algebriac equation has but one solution, which is perfectly clear.

As a matter of interest, the OP asked

In the context of simple algebriac equations
Which is why I asked if he understood differential equations.

Not all Maxwell's equations are DEs.
Can you say the same for Einstein's Field Equations?