Unit problem in differential equation

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Discussion Overview

The discussion revolves around the units involved in a differential equation, specifically focusing on the variable \( t \) and the constant \( c \). Participants explore the implications of dimensional analysis and non-dimensionalization in the context of the equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a differential equation \( \frac{dN}{dt} = c \times (\text{other terms with no unit}) \) and questions the units of \( t \) and \( t_1 \) when \( t_1 = \frac{t}{c} \).
  • Another participant suggests a non-dimensionalization approach, introducing quantities with subscripts to clarify the dimensions and transforming the equation into a non-dimensional form.
  • Some participants assert that \( t \) will be in nanoseconds (ns) based on the dimensions of the right-hand side of the equation, which also has dimensions of time-1.
  • There is a proposal to multiply \( t \) by \( c \) to create a dimensionless version of the equation, leading to a new variable \( \tau = ct \).
  • A later reply acknowledges a mistake in the initial approach and confirms the need to multiply \( t \) by \( c \) to obtain values of \( N \) for different times, noting that time would then have no unit.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of \( t \) and \( c \), particularly regarding whether to divide or multiply \( t \) by \( c \). The discussion remains unresolved as multiple approaches and interpretations are presented.

Contextual Notes

Participants rely on specific assumptions about the dimensions of the variables involved, and the discussion highlights the complexity of dimensional analysis without reaching a consensus on the best approach.

sunipa.som
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I have one differential equation like
dN/dt=c*(other terms with no unit)
unit of c is 1/ns. Now if I solve this equation, I will get value of N corresponding to t.
(1) Then what will be the unit of t?
(2) and if I calculate dN/dt1=(other terms with no unit). where t1=t/c.
Then what will be the unit of t1?
 
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As I see it you have three possible units [N], [t] and [c], and I nonlimensionalise according to [itex]N=N_{d}\bar{N}[/itex], [itex]t=t_{d}\bar{t}[/itex] and [itex]c=c_{d}\bar{c}[/itex]. There I put all the dimensions into the quantities with subscript d. So this turns the differential equation into:
[tex] \frac{d\bar{N}}{d\bar{t}}=\frac{c_{d}t_{d}}{N_{d}}\bar{c}[/tex]
Where [itex]c_{d}t_{d}/N_{d}[/itex] is a nondimensional quantity. Does this clear things up?
 
t will be in ns. Your RHS has dimensions of time-1. dN/dt also has dimensions of time-1. So everything matches.

Are you sure you want to divide t by c? Maybe I'm confused, but I think it would make more sense to multiply t by c, in order to get a dimensionless version of the equation. For instance, if you define [itex]\tau=ct[/itex], then you get [itex]\frac{dN}{d\tau}=\mbox{(other terms)}[/itex] and c has gone away.
 
pmsrw3 said:
t will be in ns. Your RHS has dimensions of time-1. dN/dt also has dimensions of time-1. So everything matches.

Are you sure you want to divide t by c? Maybe I'm confused, but I think it would make more sense to multiply t by c, in order to get a dimensionless version of the equation. For instance, if you define [itex]\tau=ct[/itex], then you get [itex]\frac{dN}{d\tau}=\mbox{(other terms)}[/itex] and c has gone away.
--------------------------
Thank you. Sorry for mistake. I have to multiply t by c. Then I will get value of N for different times but then time has no unit.
 

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