Significance of delta in expressions.

[Minki]

Can you always put numbers directly into an expression which has the delta triangle against its variables? For example Faraday's Law is always shown with the delta triangle on top and bottom yet these are dropped when you use it for calculations; which leaves me wondering why they are there in the first place, since you never find them (for example) in Ohms Law.

Thanks.

 

[Fightfish]

The $\Delta$ symbol means "change".
For instance, $\Delta v$ refers to the change in velocity $v$

 

[Minki]

Hi Fightfish, yes I am aware of that. But my question was about its significance in the use of the expression.

 

[I like Serena]

Welcome to PF, Minki!

When you refer to Faraday's Law, I presume you mean something like the following formula?
$$\mathcal{E} = -N {{\Delta\Phi} \over \Delta t}$$
In this formula $\Delta\Phi$ is shorthand for $\Phi_2 - \Phi_1$, and $\Delta t$ is shorthand for $t_2 - t_1$.
This is about the flux at two different times (that are close together).
More specifically $\Phi_1$ is the flux at time $t_1$, and $\Phi_2$ is the flux at time $t_2$.

In other words, you could also write the formula as:
$$\mathcal{E} = -N {{\Phi_2 - \Phi_1} \over t_2 - t_1}$$
The delta symbols cannot simply be dropped from any calculation.
The only case you can drop them, is if the initial time $t_1$ is zero, and if the initial flux $\Phi_1$ is also zero.
And that is only allowed, because in that case the result is the same.

 

[Minki]

Hi I like Serena, Many thanks for your very comprehensive reply!

That's exactly what I meant. So I think it would be fair to say that if an expression can be represented as a linear graph going through the origin on both x and y-axis then the deltas are not needed, as in the case of Ohms Law.

 

[aralbrec]

Hi PF, Many thanks for your very comprehensive reply!

That's exactly what I meant. So I think it would be fair to say that if an expression can be represented as a linear graph going through the origin on both x and y-axis then the deltas are not needed, as in the case of Ohms Law.

Faraday's law is actually about the instantaneous rate of change of flux.

ε = -N dψ/dt

where calculus is used. dψ/dt is the slope of the ψ vs t graph at a particular instant of time t. Δψ/Δt approximates this slope if Δt is sufficiently small and the graph isn't too curvy at the point of interest.

When you say the graph is linear, then it has a constant slope so dψ/dt = Δψ/Δt = a constant. Only when the linear graph starts at the origin can the second part of each delta equal zero. A graph can still be linear and not start at the origin.

The deltas are always needed because emf is only induced when the flux *changes*. If the flux is constant so that Δψ = 0, there is no emf, ie ε=0. If you don't write these deltas, the equation is not complete and even misleading. Writing ε = -N ψ/t is neither accurate mathematically nor does it describe to anyone reading it that ε depends on the rate of change of flux.
I was about to tell you that Ohm's Law is nothing like that but of course that's not quite true and is probably why you brought this up at all :) But in the application of Ohm's law, it is understood that the voltage is the voltage across the circuit element so ΔV = V_{at top of element} - V_{at bottom of element} is understood to mean 'V'. And in a perfectly analogous way to the flux, the voltage drop across that space is the average electric field E = - ΔV/Δx

So I know the point you are making -- we've dropped the ΔV in Ohm's law. The problem with the change in flux is dropping the Δψ/Δt would make the problems harder to solve because important information would be missing (understood to be there) that would make it difficult to solve equations with derivatives in them.

 

[Minki]

Thanks for that aralbrec. I had forgotten that the expression must represent the need for change. Without a change in flux the expression would always equal zero, which of course is not the case with Ohms Law which represents simple resistance. So I think that completes the solution to my issue.

Thanks again.