What is the Difference Between ∫X.dY and ∫Y.dX in Physical Applications?

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Discussion Overview

The discussion centers on the physical interpretation of the integrals ∫X.dY and ∫Y.dX, exploring their meanings in various physical contexts, particularly in relation to work done and changes in state variables such as force, displacement, pressure, and volume. Participants seek to clarify how these mathematical expressions translate into physical applications.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that ∫X.dY represents the area bounded by the curve and the Y-axis, while ∫Y.dX represents the area bounded by the curve and the X-axis, but struggles to find physical applications for these integrals.
  • Another participant emphasizes that X and Y are arbitrary variables and that without definitions, the integrals lack meaning.
  • A participant agrees that ∫F.dX calculates work done by integrating force over displacement, but questions the meaning of ∫X.dF, suggesting it involves displacement but not in the same way as the first integral.
  • There is a discussion about the integral ∫v.dp, with one participant expressing uncertainty about its physical interpretation, while another suggests it could relate to integrating volume for a change in pressure.
  • Participants clarify that the displacement referred to in the context of ∫X.dF is the total displacement from the origin, not a small change in displacement.
  • One participant references a thermodynamic equation involving dH, dU, and pdV, questioning the meaning of the term Vdp within that context.

Areas of Agreement / Disagreement

Participants express differing views on the physical meanings of the integrals discussed, with no consensus reached on the interpretation of ∫v.dp or the relationship between ∫X.dF and ∫F.dX. The discussion remains unresolved regarding the physical significance of these integrals.

Contextual Notes

Participants highlight the importance of defining variables to make the integrals meaningful, indicating that the discussion may depend on specific contexts or definitions that are not fully articulated.

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What is the difference between ∫X.dY and ∫Y.dX in the physical world? I know what the difference is in pure mathematics. ∫X.dY represents the are bounded by the curve and the Y axis while ∫Y.dX represents the area bounded by the curve and the X axis. But I am unable to translate this into physical situations.

For example when we are doing ∫F.dX, we are calculating force multiplied by displacement for very small displacements and then adding them all up. Similarly ∫X.dF should mean that we are multiplying the displacement of the particle for a very small change in F with that small change. Should they not mean the same thing and give the same result? Yet only the first is called the work done. One other example that I can think of is that of pressure-volume work. ∫p.dv is the work done. I can't even think of what ∫v.dp means. What does it mean?
 
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X and Y are arbitrary variables. If you do not define them, the integrals are meaningless.

I can't even think of what ∫v.dp means
Integrate the volume for a pressure change. If volume is a function of pressure, this is possible (I am not sure if there is a useful physical interpretation for that).
 
transparent said:
For example when we are doing ∫F.dX, we are calculating force multiplied by displacement for very small displacements and then adding them all up.
Correct.

Similarly ∫X.dF should mean that we are multiplying the displacement of the particle for a very small change in F with that small change.
No. we are multiplying "the displacement of the particle", not "the displacement of the particle for a very small change in F". In other words, "the displacement from origin", not "the small change in displacement".

I can't even think of what ∫v.dp means. What does it mean?
If you write down a mathematical formula "at random", usually it doesn't mean anything physically.
 
AlephZero said:
Correct.


No. we are multiplying "the displacement of the particle", not "the displacement of the particle for a very small change in F". In other words, "the displacement from origin", not "the small change in displacement".

That makes sense. But ∫v.dp should mean something since dH=dU+pdV+Vdp. Here dU is the change in internal energy and pdV is the work done. What is Vdp?
 

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