- #1

eoghan

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You have a particle moving to the left as time goes on. Now if you reverse the time the particle will move to the right. Does it mean that the system is not symmetric under time reversal?

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- Thread starter eoghan
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- #1

eoghan

- 205

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You have a particle moving to the left as time goes on. Now if you reverse the time the particle will move to the right. Does it mean that the system is not symmetric under time reversal?

- #2

The_Duck

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- #3

nnnm4

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- #4

eoghan

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On the contrary, rotation, parity, charge etc are symmetries of the specific system and not physical laws, aren't they?

- #5

betel

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On the contrary, rotation, parity, charge etc are symmetries of the specific system and not physical laws, aren't they?

It is precisely as The_Duck said. If I show you a movie of something physical process that is symmetric under time reversal you will not be able to tell me wether the movie is going forward or backward.

One always determines time symmetry for the equations of motion (or the principle they are derived from) of a system. If those equations do not change under t-> -t, then you have time symmetry and a specific motion is then determined by initial conditions.

You are mixing some stuff here. Parity and rotations are symmetries of nature (well, parity only is to some extent). From continuous symmetries we can deduce conserved quantities which in the case of rotations is angular momentum.

Charge is the conserved quantity to a complex rotation in the Lagrangian.

Another symmetry is charge conjugation, which exchanges particles and anti particles.

- #6

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This is correct.

For the situation presented by the OP the relevant law is Newton's first law. The forward time particle is initially going to the right and continues going to the right at the same speed per Newton's first law. The reverse time particle is initially going to the left and continues going to the left at the same speed per Newton's first law. Newton's first law is therefore symmetric under time reversal.

- #7

eoghan

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You are mixing some stuff here. Parity and rotations are symmetries of nature (well, parity only is to some extent). From continuous symmetries we can deduce conserved quantities which in the case of rotations is angular momentum.

But can we deduce conserved quantities from symmetries of the physics laws or we can do it just from symmetries of nature?

- #8

betel

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Whenever there is discrepancy you have to correct your theory (not blame it on nature)

That was the case with parity. For a long time everybody assumed and was sure that is an exact symmetry of nature. Until somebody really measured decays on found one favourite chirality.

So now parity is only an approximate symmetry of nature.

- #9

eoghan

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Let's suppose I have a particle on the top of a gaussian-shaped hill (i.e. a hill symmetric under rotations along vertical axis). If I rotate the system along the vertical axis I can't say that the system has been rotated, because it looks the same. So I say that the system is symmetric.

But if the particle was not on the top of the hill, but on a side, then if I rotate the system I can see that the particle is changing its position, so the system is not symmetric anymore under rotations.... is it true?

- #10

betel

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In general a specific state will not be symmetric.

But the laws (equations of motion) are. Evolution will look exactly the same whether time is running forward or backward. The only difference you will have to allow for is different initial condition.

But the point with particle on the side not being symmetric is a good point, although this will lead to some sophisticated physics. May I ask what your level of physics is. Have heard about the Higgsparticle.?

- #11

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You are correct, the second system is not axisymmetric any more. However, usually time-reversal symmetry is discussed wrt the laws and not specific systems, so I don't understand the purpose of your question here in the context of the OP.

Let's suppose I have a particle on the top of a gaussian-shaped hill (i.e. a hill symmetric under rotations along vertical axis). If I rotate the system along the vertical axis I can't say that the system has been rotated, because it looks the same. So I say that the system is symmetric.

But if the particle was not on the top of the hill, but on a side, then if I rotate the system I can see that the particle is changing its position, so the system is not symmetric anymore under rotations.... is it true?

Do you understand the difference between a differential equation and the initial conditions? The laws of physics are cast as differential equations, and a specific system is represented by the initial conditions.

- #12

eoghan

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I'm attending the third year in physics and I studied QM last semester and now I'm studying an introduction to elementary particles physics. But I've never studied symmetries deeply and now I'm a bit confused!betel said:But the point with particle on the side not being symmetric is a good point, although this will lead to some sophisticated physics. May I ask what your level of physics is. Have heard about the Higgsparticle.?

I heard about Higgs particle... but I've never understood that much

- #13

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I just have never ever heard of aOk, so a system is said to be time reversal invariant ...

- #14

betel

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I just have never ever heard of asystembeing described as time reversal invariant. I have only ever heard that description applied tolawsof physics.

That is not quite true. You usually attribute the symmetries to the Langrangian describing a system. Some quantities and their corresponding conserved quantities can only be read off from the Lagrangian and not the equations of motion, e.g. charge.

The Higgs particle utilizes the property that the general system is invariant under a cvertain symmetry but a certain state is not.

In this case the electroweak symmetry. The Lagrangian is invariant under this symmetry but the ground state (vacuum state) with the least energy does not respect this symmetry, so it is lost in everyday life, unless at very high energies (more than can be realized on earth).

- #15

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Not in my experience, at least not for time-reversal symmetry.You usually attribute the symmetries to the Langrangian describing a system.

- #16

Phrak

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You have a particle moving to the left as time goes on. Now if you reverse the time the particle will move to the right. Does it mean that the system is not symmetric under time reversal?

Yes, the system is not symmetric under time reversal. Velocity V is replaced with -V.

I would be curious (but doubtful) to see how applying a Lagrangian to a quantum system, rather than the Newtonian F=ma, could change this.

- #17

betel

- 318

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The confusion in here stems from the fact that there is difference whether one considers the symmetries of a general system or a specific state in a given system.

In other words it always depends on the background you re working in.

For simplicity consider motion in free space.

Then the motion of any number of particles is symmetric under time reversal, that means for any given configuration of initial conditions I can find another set of initial conditions that will look like the same evolution if time is running backwards.

Now if you want to consider the background of one particle already moving in one fixed way in empty space, and ask how other particles would evolve in this background, then there will be no time reversal symmetry any more as you one set particle explicitly breaks the symmetry.

- #18

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