Undergrad Check invariance under time-reversal?

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To check if the equation of motion for a particle with a given potential is invariant under time reversal, it is crucial to analyze the nature of the potential. The discussed potential includes a term with a time derivative, which indicates a frictional force and is not conservative, thus violating time-reversal symmetry. The presence of this friction term means that the potential cannot be considered a standard potential, as it depends on time derivatives rather than solely on spatial coordinates. The discussion emphasizes that only time-independent potentials are invariant under time reversal, and the inclusion of non-conservative terms complicates this analysis. Overall, the conversation highlights the importance of distinguishing between conservative potentials and those affected by friction in the context of time-reversal invariance.
FilipLand
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Hi!

How do I check if the equation of motion of the particle, with a given potential, is invariant under time reversal?

For a 2D pointlike particle with potential that is e.g $$V(x) = ae^(-x^2) + b (x^2 + y^2) +cy', where a,b,c >0$$

Can it be done by arguing rather then computing?

Thanks!
 
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Your potential seems constant ...
 
Orodruin said:
Your potential seems constant ...

How come? And what would that mean in this context? That we can tell if the particle move back or forward in time since potential is constant?
 
Something that is time independent is obviously invariant under time reversal.
 
Orodruin said:
Your potential seems constant ...

There is a y-prime in it.
 
Then it is not a potential.
 
Vanadium 50 said:
There is a y-prime in it.
yes it is
 
Orodruin said:
Then it is not a potential.

I get your point, that it's not a function of time. Thanks. But FYI it's wrong to say something is not a potential due to time independency.
 
FilipLand said:
But FYI it's wrong to say something is not a potential due to time independency.
I never said that. I said that it is not a potential because it contains a time derivative of the coordinates.
 
  • #10
Orodruin said:
I never said that. I said that it is not a potential because it contains a time derivative of the coordinates.

Why wouldn't it be? I have an example exercise where that is the potential we work with?
 
  • #11
FilipLand said:
Why wouldn't it be? I have an example exercise where that is the potential we work with?
Because the potential is a function of the coordinates only. Not of time derivatives of the coordinates. Please give more details of what you are reading.
 
  • #12
Orodruin said:
Because the potential is a function of the coordinates only. Not of time derivatives of the coordinates. Please give more details of what you are reading.

The statement was not true in this case, you can notice in which direction time flows! The time derivative term is a friction term which is not time reversible.
 
  • #13
FilipLand said:
The statement was not true in this case, you can notice in which direction time flows! The time derivative term is a friction term which is not time reversible.
A friction term is not part of any potential because friction is not conservative. It is misleading to include friction terms as "potential" terms.
 

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